Answer (A): To obtain the region swept out by the first disk, remove a square of side length $s-4$ from the center of the original square and replace the $4$ unit squares at the corners of the original square with $4$ quarter-circles of radius $1$. The area of this region is
$$s^2 - (s-4)^2 - 4 + \pi = 8s - 20 + \pi.$$
The region swept out by the second disk is the disjoint union of $4$ rectangles, each with length $s$ and width $2$, and $4$ quarter-circles of radius $2$, so its area is $8s + 4\pi$. Therefore
$$8s + 4\pi = 2(8s - 20 + \pi).$$
Solving this equation gives $s = 5 + \frac{\pi}{4}$, so the requested sum is $5 + 1 + 4 = 10$.