Solution: Choose A. Reason: Cube: $S=6 a^{2}, a=\sqrt{\frac{S}{6}}, V_{1}=a^{3}=\sqrt{\frac{S^{3}}{216}}$; Equilateral cylinder: $S=2 \pi R^{2}+$
$$
\begin{array}{l}
2 \pi R \cdot 2 R=6 \pi R^{2}, R=\sqrt{\frac{S}{6 \pi}}, V_{2}=\pi R^{2} \cdot 2 R=2 \pi \sqrt{\frac{S^{3}}{216 \pi^{3}}}=\sqrt{\frac{S^{3}}{54 \pi}}>\sqrt{\frac{S^{3}}{216}}=V_{1} \text {; Sphere: } \\
S=4 \pi R^{2}, R=\sqrt{\frac{S}{4 \pi}}, V_{3}=\frac{4}{3} \pi R^{3}=\sqrt{\frac{S^{3}}{36 \pi}}>\sqrt{\frac{S^{3}}{54 \pi}}=V_{2}, \text { therefore } V_{1}<V_{2}<V_{3} .
\end{array}
$$
Example 8 (4th "Hope Forest" Invitational Competition Question) Given that the surface areas of a cube, an equilateral cylinder, and a sphere are all $S$, their volumes are $V_{1}$, $V_{2}$, and $V_{3}$ respectively, then ( ).
A. $V_{1}<V_{2}<V_{3}$
B. $V_{3}<V_{2}<V_{1}$
C. $V_{3}<V_{1}<V_{2}$
D. $V_{2}<V_{1}<V_{3}$