Algebra needed?

Connect $BG$ . It is clear that $AF \parallel BG$ . Then the area of triangle $AFG$ is same as the area of the triangle $AFB$ , which is $\frac{3}{4} \times 24 =18$ .

If we assume, as the question implies, that the area of $\triangle AFG$ is the same for any measurements such that $\triangle ABC$ is similar to $\triangle BDE$ , $BF:FC = DG:GE = 3:1$ , and the area of $\triangle ABC = 24$ , then we can solve it for an easy specific case that satisfies those conditions, for example, $\triangle ABC \cong \triangle BDE$ , $\angle ABC = \angle BDE = 90°$ , $AB = BD = 12$ , $BC = DE = 4$ , $BF = DG = 3$ , and $FC = GE = 1$ .

Then $\triangle AFG$ has a base $FG = BD = 12$ and a height of $FB = 3$ , for an area of $A = \frac{1}{2}bh = \frac{1}{2} \cdot 12 \cdot 3 = \boxed{18}$ .