Call \(r\) the radius and \(C\) the center of the small circle.
Consider the triangle \(ABC\) with a right angle at \(B\).
We know the following lengths:
\(AB = 1 - r\)
\(BC = \dfrac{1}{2}\)
\(AC = 1 + r\)
By Pythagoras' theorem we have
\[(1-r)^2 + \left(\dfrac{1}{2}\right)^2=(1+r)^2 \]
which gives
\[r = \dfrac{1}{16}\,.\]