Solution:

Let a segment $\overline{PQ}$ in $\mathcal{S}$ have endpoints on two sides meeting at corner $A,$ and let $M$ be its midpoint. Triangle $PAQ$ is right-angled at $A$ with hypotenuse $PQ = 2,$ and the median to the hypotenuse of a right triangle is half the hypotenuse, so $AM = 1.$ Conversely every point at distance $1$ from a corner (between the two adjacent sides) is such a midpoint, so the midpoints form four quarter-circle arcs of radius $1$ centered at the corners of the square.

The region these arcs enclose is the square with the four quarter-disks removed, of area $$4 - 4 \cdot \frac{\pi}{4} = 4 - \pi \approx 0.86.$$ Therefore $100k = 86.$