Say, the radius of gyration of the body is $K$.
The mass is $m$.
The height is $h$.
The radius of the body is $R$.
According to the conservation of mechanical energy,$mgh=\frac{1}{2}m{v}^2\left(1+\frac{K^2}{R^2}\right)$ or $v^2=\left(\frac{2gh}{1+\frac{K^2}{R^2}}\right)$
Note: It is independent of the mass of the rolling body.
For a ring, $K^2=R^2$
So, the speed of the ring becomes,$v_{ring}=\sqrt{\frac{2gh}{1+1}}=\sqrt{gh}$
For a solid cylinder, $K^2=\frac{R^2}{2}$
$v_{cylinder}=\sqrt{\frac{2gh}{1+\frac{1}{2}}}=\sqrt{\frac{4gh}{3}}$
For a solid sphere, $K^2=\frac{2}{5}{R}^2$
$v_{sphere}=\sqrt{\frac{2gh}{1+\frac{2}{5}}}=\sqrt{\frac{10gh}{7}}$
Among the given three bodies, the solid sphere has the greatest and the ring has the least velocity at the bottom of the inclined plane.