Inspired by Justin Wong

Let the 2 circles be tangential at point $T$ . Draw the tangent at $T$ , and call the line $\ell$ . Let $A', B'$ be the intersection of $\ell$ with $OA$ and $OB$ respectively.

Observe that the small circle is the incircle of the equilaterial triangle $O A' B '$ , which has height equal to the radius of the circle, IE $1$ . Hence, the radius of the incircle is $\frac{1}{3}.$

This only works out because the given angle is $60 ^ \circ$ . Otherwise, it will require more work, like in Brian's solution

Let $P$ be the center of the radius $r$ circle inscribed in sector $AOB$ , and let $Q$ be the point where this circle touches $OA.$

Then $\Delta PQO$ is right-angled at $Q$ with $PQ = r, PO = 1 - r$ and $\angle QOP = \frac{\pi}{6}.$ We then have that

$\sin(\angle QOP) = \dfrac{PQ}{PO} \Longrightarrow \dfrac{1}{2} = \dfrac{r}{1 - r} \Longrightarrow 1 - r = 2r \Longrightarrow r = \dfrac{1}{3} = \boxed{0.333}$

to $3$ decimal places.