You only made one computational error and a notational error.
Here is your solution (at least the first part that follows) re-written:
You are minimizing the cost of the fence, so you want to minimize the perimeter of the enclosure. The perimeter $P$ is $\tag{1} P=2l+3w. $ where $l$ is the length and $w$ is the width of the enclosure. It is assumed the division of the original fence is made across a width (hence the $3w$).
We want to write $P$ in terms of one variable. For this we use the given information that the area is $1.5\cdot10^6\,\rm ft^2$. So $lw= 1.5\cdot10^6\,\rm ft^2,$ or, $\tag{2} l={1.5\cdot10^6\over w} $ Substituting $(2)$ into $(1)$ gives $ P=2\bigl({1.5\cdot10^6\over w}\bigr) +3w $ or $ P(w)= {3\cdot10^6\over w} +3w $ (this is where you made your mistake).
We also need to find the appropriate range of values for $w$. Here, $w$ could be possibly any positive, finite length.
So, you want to find the minimum value of $P$ over the interval $(0,\infty)$.
Towards this end, evaluate P'(w)= - {3\cdot10^6\over w^2} +3. Set P'(w)=0 and find the solutions in the interval $(0,\infty)$: $ - {3\cdot10^6\over w^2} +3 = 0 \quad\iff\quad {3\cdot10^6 } =3w^2\quad\iff\quad w=10^3. $
So $w=1000\,\rm ft$ gives the only critical point of $P$ in the interval $(0,\infty)$. We need to examine what goes on near the endpoints of our interval before declaring that $1000\,\rm ft$ gives the answer. As you can easily convince yourself $\lim\limits_{w\rightarrow 0^+} P(w)$ and $\lim\limits_{w\rightarrow \infty} P(w)$ are both infinite. So the cost of the enclosure is minimized when $w=1000\,\rm ft$.
And to explicitly answer the question (don't forget to do this): "he can do this by taking the width to be $1000\,\rm ft$, the length to be ${1.5\cdot 10^6\over1000}=1500\,\rm ft$ and spliting the enclosure by constructing another width".