[Solution] Connect $D E$. Given $S_{\triangle A B E}=S_{B E F D}$,  
we have $S_{\triangle A D E}=S_{\triangle F D E}$, thus $E D \parallel A C$.  
Also, $A D: D B=2: 3$,

then
$S_{\triangle A D E}: S_{\triangle E D B}=2: 3$.
Since $E D \parallel A C$, we have

thus
$$
S_{\triangle E D B}: S_{\triangle A B C}=3^{2}:(3+2)^{2}=9: 25 \text {, }
$$

and
therefore $S_{\triangle A B E}=S_{\triangle E D B}+S_{\triangle A D E}=\frac{18}{5}+\frac{12}{5}=6$.
Hence, the answer is $(C)$.
$17 \cdot 147$ As shown in the figure, the area of $\triangle ABC$ is 10. Points $D, E, F$ are three points on $AB, BC, AC$ respectively, different from $A, B, C$, and $AD=2, DB=3$. If the area of $\triangle ABE$ and quadrilateral $DBEF$ are equal, then the area is
(A) 4.
(B) 5.
(C) 6.
(D) $\frac{5}{3} \sqrt{10}$.
(E) cannot be uniquely determined.