let's say x is the side of the square.

The radius of the inner circle is x/2

The radius of the outer circle is x*sqrt(2)/2.

The difference of the radius is x*sqrt(2)/2-x/2=a


x/2*(sqrt(2)-1)=a

x=2a/(sqrt(2)-1)

Let's say
R=radius of the big circle
r=radius of the small circle
a=R-r according to the question

Look at the draw and take half of the square, you have a rectangular triangle and we can use pyhtagore theorem. You can see that 2r is one side of the square and that 2R is the hypothenuse of the square.

(2R)^2 = (2r)^2 + (2r)^2
R=r*sqrt(2)

Now let's use R-r=a
a+r=r*sqrt(2)
a/(sqrt(2) -1)=r

We want one side, which is 2r so:
2r= 2a/(sqrt(2) -1)