For a largest fitted square, each side must touch an inner corner of one of the removed $1$-inch squares; otherwise the fitted square could be enlarged. Thus it surrounds the central $3\times3$ square and adds four congruent right triangles, one along each side.

The central square has area $3\cdot3=9$. Each added triangle has base $3$ and height $1$, so the four triangles have total area $$4\left(\dfrac{3\cdot1}{2}\right)=6.$$

The fitted square has area $9+6=15$.

Thus, C is the correct answer.