Solution:

Assume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller area).

If two triangles have their bases on the same line and the same opposite vertex, then it follows from the $\frac{1}{2} b h$ formula that their areas are in the same ratio as their bases. In particular (brackets denote areas),
$$\frac{[D B G]}{[A B C]} = \frac{[D B G]}{[A B G]} \cdot \frac{[A B G]}{[A B C]} = \frac{D B}{A B} \cdot \frac{B G}{B C} = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9},$$
and similarly $[E B F] / [A B C] = 1 / 9$. Subtracting gives $[D E F G] / [A B C] = 1 / 3$, so the answer is $[A B C] / 3 = 17$.