jmc

We first find the length of line segment $FG$. Since $DC$ has length $6$ and $DF$ and $GC$ have lengths $1$ and $2$ respectively, $FG$ must have length $3$. Next, we notice that $DC$ and $AB$ are parallel so $\angle EFG\cong \angle EAB$ because they are corresponding angles. Similarly, $\angle EGF\cong \angle EBA$. Now that we have two pairs of congruent angles, we know that $△FEG\sim △AEB$ by Angle-Angle Similarity.

Because the two triangles are similar, we have that the ratio of the altitudes of $△FEG$ to $△AEB$ equals the ratio of the bases. $FG:AB=3:6=1:2$, so the the ratio of the altitude of $△FEG$ to that of $△AEB$ is also $1:2$. Thus, the height of the rectangle $ABCD$ must be half of the altitude of $△AEB$. Since the height of rectangle $ABCD$ is $3$, the altitude of $△AEB$ must be $6$. Now that we know that the base and altitude of $△AEB$ are both $6$, we know that the area of triangle $AEB$ is equal to $\frac{1}{2}$base $\times$ height $=(\frac{1}{2})(6)(6)=\boxed{18}$ square units.