\section*{18. (12
Let $CF = a$. In $\triangle AFM$, $AM = \sqrt{3}$, $FM = \sqrt{a^2 + 1}$, $AF = \sqrt{a^2 + 4}$.
Then $\frac{1}{2} \times \sqrt{3} \times \sqrt{a^2 + 1} = \frac{1}{2} \times \sqrt{a^2 + 4} \times \frac{\sqrt{30}}{5}$. Solving, we get $a = 1$.
Thus $V_{C-AFM} = V_{F-ACM} = \frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{6}$.