The threshold frequency for a metallic surface is nu0. If light of frequency 2nu0 is incident on this surface, the maximum velocity of photoelectrons…

Maximum kinetic energy $K_{\text{max}} = \frac{1}{2}m v^2 = h(\nu - \nu_0)$. For frequency $2\nu_0$: $\frac{1}{2}m v_1^2 = h(2\nu_0 - \nu_0) = h\nu_0$. For frequency $5\nu_0$: $\frac{1}{2}m v_2^2 = h(5\nu_0 - \nu_0) = 4h\nu_0$. Taking the ratio: $\frac{v_1^2}{v_2^2} = \frac{1}{4} \implies \frac{v_1}{v_2} = \frac{1}{2}$.