In pure rolling motion of a disc, the total kinetic energy ($K_{total}$) is the sum of its translational kinetic energy ($K_t$) and rotational kinetic energy ($K_r$). $K_t = \frac{1}{2}mv^2$ $K_r = \frac{1}{2}I\omega^2$ For a disc, the moment of inertia about its central axis is $I = \frac{1}{2}mR^2$. Also, $v = R\omega$ for pure rolling. So, $K_r = \frac{1}{2} \left(\frac{1}{2}mR^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4}mv^2$ $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ The ratio of total kinetic energy to rotational kinetic energy is: $\frac{K_{total}}{K_r} = \frac{\frac{3}{4}mv^2}{\frac{1}{4}mv^2} = \frac{3}{1} = 3:1$