When light of frequency twice the threshold frequency is incident...

Maximum kinetic energy is expressed as {K}_\max = hf - ϕ, here, $ϕ$ is work function of metal.

For first case, $\dfrac{1}{2}{m{v}_{1}}^{2} = h\left[\left(2{f}_{0}\right) - {f}_{0}\right]$

For second case, $\dfrac{1}{2}{m{v}_{2}}^{2} = h\left[\left(5{f}_{0}\right) - {f}_{0}\right]$

$⇒ \dfrac{{{v}_{2}}^{2}}{{{v}_{1}}^{2}} = \dfrac{4}{1} ⇒ \dfrac{{v}_{2}}{{v}_{1}} = 2$