We know that the triangles are congruent. The circles are congruent. And EFGH is a square. I don't even know where to begin with this one besides to say that angle ADE is equal to 90-m(EAD), but I feel like it's probably asking for something more specific.
O, the center of the central triangle, must be at the center of the square. so the angle $ O\hat AB$ is $\frac{\pi}4$. using elementary properties of tangents this gives $B \hat AF = \frac{\pi}6$
if we denote by $M$ the midpoint of $EF$ and let $x$ measure $AM$, whilst $r$ is the radius of a circle, then: $$ x^2 + r^2 = \frac12 $$ and $$ \frac{x-r}{x+r} = \sin \frac{\pi}6 = \frac12 $$ which after simplification gives: $$ r = \frac1{2\sqrt{5}} $$