When a certain metallic surface is illuminated with monochromatic...

Using photoelectric equation, $hf - h{f}_{0} = \dfrac{1}{2}m{v}^{2} = e{V}_{s}$
Or \left(\dfrac{hc}{\lambda} - \dfrac{hc}{{\lambda}_{0}}\right) = e{V}_{s}$<br />For the first case,$\dfrac{hc}{\lambda} - \dfrac{hc}{{\lambda}_{0}} = e\left(3{V}_{0}\right) ...(i)
For the second case, \dfrac{hc}{2\lambda} -\dfrac{hc}{{\lambda}_{0}} = e\left({V}_{0}\right) ...(ii)
Solving {\lambda}_{0} = 4\lambda