jmc

In triangle $ABF$, the two acute angles are equal since $AB=AF$. Also, the measure of $\angle A$ is $180^{\circ }(6-2)/6=120^{\circ }$. Letting $x$ be the measure of $\angle ABF$, we have $$120^{\circ }+x+x=180^{\circ }⟹x=\boxed{30}\text{ degrees}.$$