Here is a tentative solution. Thank you copper.hat for the very helpful hint. There is one interesting thing to note here. The text suggests applying Lemma 1.1 which proves that a finite decomposition of a rectangle into sub rectangles (or cubes) has additive volume if the decomposition is almost disjoint (and sub additive in general by extension). I had difficulty extending this to countable decompositions, so I applied copper.hat's suggestion (second lemma that I prove in the solution), and an auxiliary lemma proved in the book that says the (cubic) exterior measure of a set which can be decomposed into almost disjoint closed cubes has additive measure. Obviously if I could apply the book's hint, I would not need to pass from the rectangular measure to the cubic measure, and could do the computation at the end more directly. But in any case, it's not really much different either way.
Problem Statement: If $E\subset\mathbb{R}^{d}$, then $m_{\star}(E) = m_{\star}^{R}(e)$.
PROOF: It is clear $m_{\star}(E)\geq m_{\star}^{R}(E)$, since any cover of $E$ by cubes is also a cover of $E$ by rectangles. To show the reverse inequality, we need two lemmas.
LEMMA #1: Every closed rectangle can be decomposed into a countable union of almost disjoint closed cubes.
PROOF: Obviously this is true for $\mathbb{R}^{1}$. To prove this in general, let $R = I_{1}\times\ldots\times I_{d}$ be any cube in $\mathbb{R}^{d}$ ($d\geq2$). Take the smallest interval (call it $I_{0}$ say) and construct closed cubes of side length $I_{0}$, so that each has a volume of $\left|I_{0}\right|^{d}$. For each interval composing $R$, compute the value $N_{j}=\frac{|I_{j}|}{|I_{0}}$, and discard the remainder (note that for the minimal interval, the quotient will be exactly $1$). Then precisely $N_{j}$ cubes can be exactly inscribed along each $I_{j}$, and this embedding of cubes will be pairwise disjoint (it helps to draw a picture in $\mathbb{R}^{3}$ to get the general idea; a picture in $\mathbb{R}^{2}$ tends to obscure the general situation). Note that the total number of cubes embedded is $\sum_{j=1}^{d}N_{j}$, which is finite. What remains is exactly $(d)$ rectangles and the union of these remaining rectangles unioned with the union of the embedded cubes gives back $R$ exactly. (Note that some of the $d$ remaining rectangles may have $0$ volume, since we may have $|I_{j}|=k|I_{0}|$ for some positive integer k; in fact, this occurs for at least one interval, namely the smallest interval). We can continue this construction indefinitely, applying the procedure to each of the remaining rectangles at each step of the decomposition (note that the volumes of the remaining rectangles goes to zero in the limit). Since the total number of cubes at the end of the construction is a countable union of a countable union of finite unions of cubes (mouthfull), the total number of cubes embedded is countable. By construction, they are also all almost disjoint from each other. Finally, let $Q=\bigcup_{k=1}^{\infty}$ where $\{Q_{k}\}_{k=1}^{\infty}$ is some enumeration of the embedded cubes, and put $x\in Q$. Then clearly, $x\in R$ as well, since each embedded cube is completely contained in $R$. Now put $x\in R$. Then $x\in Q_{k}$ for some index $k$ since it is contained in some subrectangle of $R$ (whether on the boundary of $R$ or the interior, it does not matter since each $Q_{k}$ is closed), which after going a large enough number of steps in the construction, is eventually contained in some $Q_{k}$. Therefore, $R=\cup_{k=1}^{\infty}Q_{k}$. (Note: there are probably more elgant ways to prove this; a geometric-series-ish construction comes to mind). QED.
LEMMA #2: If $R\subset\mathbb{R}^{d}$ is any closed rectangle (or closed cube), then $m_{\star}^{R}(R)=|R|=m_{\star}(R)$.
PROOF: The RHS equivalence follows from the computations given in the text (in particular, example 4). As for the LHS equivalence, note immediately that $m_{\star}^{R}(R)\leq|R|$ since $R$ covers itself. To prove the reverse inequality, we apply the "$\epsilon$-fattening" trick as in example 2. Let $\epsilon>0$ be given and suppose $\{R_{j}\}_{j=1}^{\infty}$ is a cover of $R$ by closed rectangles. Let $\{S_{j}\}_{j=1}^{\infty}$ be a collection of open rectangles such that for each $j=1,2,\ldots$ we have that $|S_{j}|\leq(1+\epsilon)|R_{j}$ and each $S_{j}\supset R_{j}$. (Note that the indexing sets coincide). Then it is clear $R\subset\bigcup_{j=1}^{\infty}S_{j}$, and since $R$ is compact, we can extract a finite number of $S_{j}$ such that $R\subset\bigcup_{j=1}^{N}S_{j}$. Consequently, \begin{equation*} |R|\leq\sum_{j=1}^{N}|S_{j}|\leq(1+\epsilon)\sum_{j=1}^{N}|R_{j}|\leq(1+\epsilon)\sum_{j=1}^{\infty}|R_{j}|. \end{equation*} Since both $\epsilon$ and the cover $\{R_{j}\}_{j=1}^{\infty}$ were arbitrary, we conclude $|R|\leq m_{\star}^{R}(R)$ as desired. QED
Returning to our proof that $m_{\star}(E)\leq m_{\star}^{R}(E)$ (and thus proving they are equal), choose any $\epsilon>0$ and let $\{R_{j}\}_{j=1}^{\infty}$ be any cover of $E$ by closed rectangles such that $m_{\star}^{R}(E)\geq\sum_{j=1}^{\infty}|R_{j}|-\epsilon$. Then by the first lemma, $R_{j}=\cup_{k=1}^{\infty}Q_{k}^{j}$ for each $j=1,2,\ldots$ This means $E\subset\bigcup_{j=1}^{\infty}R_{j}=\bigcup_{j=1}^{\infty}\bigcup_{k=1}^{\infty}Q_{k}^{j}$. And so, we compute: \begin{align*} m_{\star}^{R}(E)+\epsilon &\geq \sum\limits_{j=1}^{\infty}|R_{j}| &\text{Consequence of definition of outer rectangular measure.}\\ &= \sum\limits_{j=1}^{\infty}m_{\star}(R_{j}) &\text{Consequence of the second lemma.} \\ &= \sum\limits_{j=1}^{\infty}m_{\star}\left(\bigcup_{k=1}^{\infty}Q_{k}^{j}\right) &\text{Consequence of the first lemma.} \\ &= \sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty}|Q_{k}^{j}| &\text{Consequence of property 5 of outer cubic measure from text}\\ &= \sum\limits_{n=1}^{\infty}|Q_{n}^{\star}| &\text{Just notation to collapse the double sum into a single sum} \\ &\geq \inf\limits_{\{Q_{j}\}_{j=1}^{\infty}\supset E}\sum\limits_{j=1}^{\infty}|Q_{j}|\\ &= m_{\star}(E). \end{align*} Since $\epsilon$ is arbitrary, the conclusion follows and $m_{\star}^{R}=m_{\star}(E)$. QED.
NOTE: The "computations from the text" refer to proofs that $m_{\star}(R) = |R|$ and $m_{\star}(Q) = |Q|$ for both open and closed cubes and rectangles. I only need to prove $m_{\star}^{R}(R) = |R|$ for closed rectangles (and consequently cubes) for purposes of the problem. Finally, "property 5 from the text" refers to the fact that if a set is decomposable into an almost disjoint union of cubes, its outer measure becomes additive.