[Solution] Given $F B^{\prime} \perp A C$, and $\angle C=60^{\circ}$, then $\triangle F B^{\prime} C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Let $B^{\prime} C=x$, then $F C=2 x$, thus $F B^{\prime}=\sqrt{3} x$.
Also, $B F=B^{\prime} F=\sqrt{3} x$.
Construct $F D \perp A B$, then by $\angle B=60^{\circ}$, we know
$$
\begin{aligned}
& B D=\frac{\sqrt{3}}{2} x . \quad D F=\frac{3 x}{2} . \\
\because \quad & \angle B F E=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}, \\
\therefore \quad & \angle D F E=45^{\circ} . \\
\therefore \quad & D E=D F=\frac{3 x}{2} .
\end{aligned}
$$

And $\sqrt{3} x+2 x=1$, which means $x=\frac{1}{2+\sqrt{3}}$.
Therefore, $S_{\triangle B E F}=\frac{1}{2} \cdot B E \cdot D F=\frac{1}{2} \cdot\left(\frac{\sqrt{3}}{2}+\frac{3}{2}\right) x \cdot \frac{3}{2} x$.
$$
\begin{aligned}
\therefore \quad S_{A E F C} & =S_{\triangle A B C}-S_{\triangle B E F} \\
& =\frac{\sqrt{3}}{4}-\frac{3}{4} \cdot \frac{3+\sqrt{3}}{2} \cdot\left(\frac{1}{2+\sqrt{3}}\right)^{2} \\
& =\frac{1}{8}(17 \sqrt{3}-27) .
\end{aligned}
$$

Therefore, the answer is $(C)$.
$21 \cdot 15$ As shown in the figure, an equilateral $\triangle ABC$ with side length 1, is folded along $EF$ so that $B$ lands on a point $B'$ on $AC$, and $FB' \perp AC$. Then the area of the quadrilateral $AEFC$ formed by the fold is
(A) $\frac{1}{6}(17 \sqrt{3}-27)$.
(B) $\frac{1}{6}(15 \sqrt{3}-27)$.
(C) $\frac{1}{8}(17 \sqrt{3}-27)$.
(D) $\frac{1}{8}(15 \sqrt{3}-27)$.