Energy of the incident photons on the photosensitive metal surface is 3W and then 5W where 'W' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

Using Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K = E - W$, where $E$ is the energy of the incident photons and $W$ is the work function of the metal.

For the first case, the energy of the incident photons is $E_1 = 3W$.
The maximum kinetic energy is $K_1 = 3W - W = 2W$.

For the second case, the energy of the incident photons is $E_2 = 5W$.
The maximum kinetic energy is $K_2 = 5W - W = 4W$.

The kinetic energy is related to the velocity by $K = \dfrac{1}{2}mv^2$, which implies $v \propto \sqrt{K}$.

The ratio of the velocities of the emitted photoelectrons is:
$\dfrac{v_1}{v_2} = \sqrt{\dfrac{K_1}{K_2}} = \sqrt{\dfrac{2W}{4W}} = \sqrt{\dfrac{1}{2}} = \dfrac{1}{\sqrt{2}}$

The ratio is $1 : \sqrt{2}$.