3. If the volume of a certain cylinder is numerically equal to its surface area, then the minimum possible value of the volume of the cylinder is ( ).
(A) $48 \pi$
(B) $50 \pi$
(C) $54 \pi$
(D) $66 \pi$
3.C.

Let the radius and height of the cylinder be $r$ and $h$, respectively. According to the problem, $\pi r^{2} h=2 \pi r^{2}+2 \pi r h$,
which simplifies to $h=\frac{2 r}{r-2}$.
Thus, $V=\pi r^{2} h=2 \pi \cdot \frac{r^{3}}{r-2}$.
Let $r-2=x^{2} (x>0)$, then
$$
\begin{array}{l}
V=2 \pi \cdot \frac{(2+x)^{3}}{x}=2 \pi \cdot \frac{(1+1+x)^{3}}{x} \\
\geqslant 2 \pi \cdot \frac{(3 \sqrt[3]{1 \times 1 \times x})^{3}}{x}=54 \pi,
\end{array}
$$

with equality holding if and only if $x=1$, i.e., $r=3$.
Therefore, $V_{\min }=54 \pi$.