jmc

Let the side length of the square be $s$, so the area of the square is $s^2$.

By the Pythagorean Theorem, we have $s^2+s^2=5^2$, so $2s^2=25$ and $s^2=\frac{25}{2}$, so the area of the square is $\frac{25}{2}=12.5$.

Since the diameter of the circle is $5$, its radius is $\frac{5}{2}$, and its area is $\pi {\left(\frac{5}{2}\right)}^2=\frac{25}{4}\pi$, which is approximately $19.63$.

The difference between the two areas is approximately $19.63-12.5=7.13$, which, to the nearest tenth, is $7.1$. Thus the area of the circle is greater than the area of the square by $\boxed{7.1}$ square inches.