smc

Draw rectangle $ABCD$ with $AB$ = $5$ and $BC$ = $3$. Next draw a height from vertex $B$ to diagonal $AC$ intersecting at $G$ and similarly from $D$ to $AC$ intersecting at $J$. Let's call this height $h$ and we can notice that $BG$ = $DJ$ = $h$. Since the points $E$ and $F$ are trisecting diagonal $AC$ the bases and heights of triangles $BEA$, $BEF$, $BEC$, $DEA$, $DEF$, and $DEC$ have the same areas. Hence the area of one of these triangles such as $BEF$ is $\frac{5\cdot 3}{2}$ = $\frac{5}{2}$. So our answer is $\boxed{\frac{5}{2}}$.