Solution:

The optimal configuration is when the two ends $X$ and $Y$ of the semicircle lie on the arc of the quarter circle. Let $O$ and $P$ be the centers of the quarter circle and semicircle, respectively. Also, let $M$ and $N$ be the points where the semicircle is tangent to the radii of the quarter circle.

Let $r$ be the radius of the semicircle. Since $PM = PN$, $PMON$ is a square and $OP = \sqrt{2} r$. By the Pythagorean theorem on triangle $OPX$, $1 = 2r^{2} + r^{2}$, so $r = 1/\sqrt{3}$. The area of the semicircle is therefore $\frac{\pi}{2} \cdot \frac{1}{3} = \frac{\pi}{6}$.

Therefore,
$$\frac{120A}{\pi} = \frac{120 \cdot \frac{\pi}{6}}{\pi} = 20.$$