Solution:

Let us assume all sides are of side length $3$. Consider the triangle $A_{1} A_{4} A_{5}$. Let $P$ be the point of intersection of $A_{1} A_{5}$ with $A_{4} A_{8}$. This is a vertex of the inner hexagon. Then $\angle A_{4} A_{1} A_{5} = \angle A_{5} A_{4} P$, by symmetry. It follows that $A_{1} A_{4} A_{5} \sim A_{4} P A_{5}$. Also, $\angle A_{1} A_{4} A_{5} = 120^{\circ}$, so by the Law of Cosines $A_{1} A_{5} = \sqrt{13}$. It follows that $P A_{5} = \left(A_{4} A_{5}\right) \cdot \left(A_{4} A_{5}\right) / \left(A_{1} A_{5}\right) = 1 / \sqrt{13}$. Let $Q$ be the intersection of $A_{1} A_{5}$ and $A_{16} A_{2}$. By similar reasoning, $A_{1} Q = 3 / \sqrt{13}$, so $P Q = A_{1} A_{5} - A_{1} Q - P A_{5} = 9 / \sqrt{13}$. By symmetry, the inner region is a regular hexagon with side length $9 / \sqrt{13}$. Hence the ratio of the area of the smaller to larger hexagon is $(3 / \sqrt{13})^{2} = 9 / 13$.