For a body rolling along a level surface, the translational and rotational K.E are equal. The body is

Let's assume that the body's mass is provided as m, its radius is given as R, and its velocity is given as v. The body's moment of inertia is also given as I.
A rolling item has both translational and rotational kinetic energy.
Now that it has been established that the body is rolling without slipping, the velocity is given as  $\mathrm{v}=\mathrm{R\omega }$.
As a result, the translational kinetic energy of a body is now equal to $\frac{1}{2}{\mathrm{mv}}^2.$

And the rotational kinetic energy is

$\frac{1}{2}{\mathrm{I\omega }}^2=\frac{{\mathrm{Iv}}^2}{2{\mathrm{R}}^2} \left[\because \mathrm{\omega }=\frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}\right]$

Now, using that energy ratio, we will arrive at our conclusion.

$\frac{{\mathrm{Iv}}^2}{2{\mathrm{R}}^2}=\frac{1}{2}{\mathrm{mv}}^2$

$\Rightarrow \mathrm{I}={\mathrm{mR}}^2$

Therefore that body might be a ring