Translational Kinetic Energy $=\frac{1}{2}m{v}^2$
Rotational Kinetic Energy $=\frac{1}{2}I{\omega }^2$

Translational Kinetic Energy = Rotational Kinetic Energy
$\frac{1}{2}m{v}^2=\frac{1}{2}I{\omega }^2$

Put $v=r\omega$, you will get $I=M{R}^2$ which is inertia of ring.