Ans: B
Let $x\text{ cm}$ be the radius of the upper base of the frustum. Hence, we have
$\begin{array}{rcl}
\dfrac{4}{12} & = & \dfrac{x}{9} \\
\dfrac{1}{3} & = & \dfrac{x}{9} \\
x & = & 3 \\
\end{array}$
Therefore, the volume of the frustum
$\begin{array}{cl}
= & \dfrac{1}{3} \pi (9)^2 (12) – \dfrac{1}{3} \pi (3)^2 (4) \\
= & 312 \pi \text{ cm}^2
\end{array}$
