Let's call the set $X$. The Lebesgue measure $m$ is inner and outer-regular, so in order to show $m(X)=0$ we need only show that any compact subset $C$ of $X$ has measure $0$, which in turn we can do by showing that for any $\epsilon>0$ there is an open set $U\supset C$ with $m(U)<\epsilon$.
Since $C$ is compact, it must be bounded, thus we have upper bounds for $|x|$ and $|z|$. Let's denote these $M_x$ and $M_z$. Note that $C\subseteq C'=\bigcup\limits_{z=-M_z}^{M_z} [-M_x,M_x]\times [0,10]\times \{z\}$ and since $m$ is monotonic $m(C)\leq m(C')$. To see that $m(C')=0$, let $U_n=\bigcup\limits_{z=-M_z}^{M_z} (-M_x-1,M_x+1)\times (-1,11)\times (z-1/n,z+1/n)$ which is open and a finite union of rectangles, so its measure is easy to calculate. Specifically, $m(U_n)=(2M_x+2)\times12\times 2/n$, which goes to $0$ as $n\to \infty$. Thus $m(C')=0$ so $m(C)=0$ for any compact set $C\subseteq X$, hence $m(X)=0$.