Beauty of Geometry - 1 (Easier version)

Let the radius of the bigger and smaller circle be $\boldsymbol R$ & $\boldsymbol r$ respectively.

Now, draw a line from the center of the bigger circle to that particular corner of the square which will go through the center of the smaller circle . Then draw 2 more squares as shown below:

As the diagonal of a square is equal to $\sqrt2$ times it's edge, we get, $\boldsymbol {OA+AB+BC=OC}$ $\boldsymbol {\Rightarrow R+r+\sqrt2 r =\sqrt2 R}$ $\boldsymbol {\Rightarrow r=\frac{R(\sqrt2-1)}{\sqrt2+1}}$ $\boldsymbol {\Rightarrow r=R(3-2\sqrt2)}$

So the radius of the smaller circle will be, $\boldsymbol {r=15(3-2\sqrt2)=\color{#69047E} {\boxed{2.574}}}$