A solid cylinder and a solid sphere, having same mass M and radius...

During pure rolling, the work done by the friction is zero. Therefore, we can apply conservation of mechanical energy.

$mgh = \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}I{\left(\dfrac{v}{R}\right)}^{2}$

$⇒ mgh = \dfrac{1}{2}\left(m + \dfrac{I}{{R}^{2}}\right){v}^{2}$

$⇒ v = \sqrt{\dfrac{2gh}{1 + \dfrac{I}{m{R}^{2}}}}$

For a solid cylinder $\dfrac{I}{m{R}^{2}} = \dfrac{1}{2}$ and for a solid sphere, $\dfrac{I}{m{R}^{2}} = \dfrac{2}{5}$

$\dfrac{{v}_{Cyl}}{{v}_{Sph}} = \sqrt{\dfrac{1 + \dfrac{2}{5}}{1 + \dfrac{1}{2}}}$$= \sqrt{\dfrac{14}{15}}$