Let the mass of the body be $M$ and its radius be $R$. The translational kinetic energy is given by $K_{trans} = \frac{1}{2}Mv^2$. The rotational kinetic energy is given by $K_{rot} = \frac{1}{2}I\omega^2$. For a body rolling without slipping, the velocity $v = R\omega$, so $\omega = \frac{v}{R}$. Substituting this into the rotational kinetic energy equation gives $K_{rot} = \frac{1}{2}I\left(\frac{v}{R}\right)^2$. According to the question, the rotational kinetic energy is equal to the translational kinetic energy: $K_{rot} = K_{trans}$ $\frac{1}{2}I\frac{v^2}{R^2} = \frac{1}{2}Mv^2$ Solving for $I$, we get: $I = MR^2$ This is the moment of inertia of a thin circular ring about its central axis perpendicular to its plane. Therefore, the body is a ring.