Ans: B
Let $z=kx^2\sqrt[3]{y}$, where $k\neq 0$.
When $x=12$ and $y=64$, $z=36$. We have
$\begin{array}{rcl}
36 & = & k(12)^2\sqrt[3]{64} \\
36 & = & 576k \\
k & = & \dfrac{1}{16}
\end{array}$
When $x=16$ and $y=729$, we have
$\begin{array}{rcl}
z & = & \dfrac{1}{16} (16)^2 \sqrt[3]{729} \\
z & = & 144
\end{array}$
