Two bodies, a ring and a solid cylinder of same material are...

In both, cases is about point of contact ring,

$mgh = \dfrac{1}{2}I{\omega}^{2}$

$mgh = \dfrac{1}{2}\left(2m{R}^{2}\right){V}_{R}^{2}{R}^{2}$

${V}_{R} = \sqrt{gh}$

Solid cylinder

$mgh = \dfrac{1}{2}I{\omega}^{2}$

$mgh = \dfrac{1}{2}\left(\dfrac{3}{2}m{R}^{2}\right){v}_{C}^{2}{R}^{2}$

${v}_{C} = \sqrt{\dfrac{4gh}{3}}$

$\dfrac{{v}_{R}}{{v}_{C}} = \dfrac{\sqrt{3}}{2}$