Answer
The ratio of the cube's speed to the marble's speed is $1.18$
Work Step by Step
Both the marble and the cube have $v_0=0$ and $h_0=0$. The marble has $\omega_0=0$
1) The marble (having rotational KE):
According to the principle of energy conservation: $$E_f-E_0=0$$ $$\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2+mg(-h_f)=0$$
The marble is a sphere rotating around its center, so $I=\frac{2}{5}mr^2$. Also, $\omega_f=\frac{v_f}{r}$ $$\frac{1}{2}mv_f^2+\frac{1}{5}mr^2\Big(\frac{v_f^2}{r^2}\Big)+mg(-h_f)=0$$ $$\frac{1}{2}v_f^2+\frac{1}{5}v_f^2=gh_f$$ $$v_f=\sqrt{\frac{gh_f}{\frac{1}{2}+\frac{1}{5}}}=\sqrt{\frac{10gh_f}{7}}$$
2) The cube (not having rotational KE):
According to the principle of energy conservation: $$E_f-E_0=0$$ $$\frac{1}{2}mv_f^2+mg(-h_f)=0$$ $$\frac{1}{2}v_f^2=gh_f$$ $$v_f=\sqrt{2gh_f}$$
So the ratio of the cube's speed to the marble's speed is $$\frac{\sqrt{2gh_f}}{\sqrt{\frac{10gh_f}{7}}}=\sqrt{\frac{2}{\frac{10}{7}}}=\sqrt{1.4}=1.18$$
