Solution: Choose D. Reason: Using estimation, take the right triangular prism $BCF-$
$$
\begin{array}{l}
A G D, \text{ then } V=V_{B E F-A G D}-V_{E-A G D}=S_{\triangle B C F} \cdot|A B|- \\
\frac{1}{3} S_{\triangle A G D} \cdot|E G|=\frac{1}{2} \cdot 2 \cdot 3 \cdot 3-\frac{1}{3} \cdot \frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{3}{2}= \\
9-\frac{3}{2}=\frac{15}{2} .
\end{array}
$$

Or directly find $V_{E-A B C D}=\frac{1}{3} \cdot 3^{2} \cdot 2=6$, then it can be determined that the volume of the required polyhedron must be greater than 6, only $\mathrm{D}$ fits.
Example 18 (1999 National College Entrance Examination Question) As shown in Figure 7-16, in the polyhedron $A B C D E F$, it is known that face $A B C D$ is a square with a side length of 3, $E F / / A B, E F=\frac{3}{2}, E F$ is 2 units away from plane $A C$. Then the volume of the polyhedron is
A. $\frac{9}{2}$
B. 5
C. 6
D. $\frac{15}{2}$