Solution

The ring is cut from the base of a hollow cone of height h and semi-angle$\alpha \ll 1$. Its radius is approximately $R = h \tan\alpha \approx h\alpha.$
When placed with the wide part facing the beam, light hits the inner conical surface parallel to the axis.

Each ray arrives with an angle of incidence$\alpha$ with respect to the normal to the surface. Reflection deflects the ray by an angle$2\alpha$towards the axis. The distance f from the point of incidence to the focus on the axis satisfies$R = f \tan 2\alpha$
Therefore,

$f = \frac{R}{\tan 2\alpha} = \frac{h\tan\alpha}{\tan 2\alpha} = \frac{h}{2}\bigl(1 - \tan^2\alpha\bigr)$

Since the angle at the vertex is small, $\tan^2\alpha \ll 1$, and the focus is located at a distance

$\boxed{\dfrac{h}{2}}$

from the plane of the ring.