Let $K_1$   and   $K_2$  be the maximum kinetic energies of photoelectrons emitted when two monochromatic  beams of wavelength  ${\lambda }_1$  and ${\lambda }_2$  , respectively are incident on a metallic surface. If  ${\lambda }_1=3{\lambda }_2$  then 

using einstein’s photoelectric equation.
 ${\left(KE\right)}_{\max }=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
where ${\lambda }_0$,  = cur off wavelength.
so,  $k_1=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}$
 $\Rightarrow 3k_1=\frac{3hc}{{\lambda }_1}-\frac{3hc}{{\lambda }_0}$
 $\Rightarrow 3k_1=\frac{hc}{{\lambda }_2}-\frac{3hc}{{\lambda }_0}\mathrm{....}\left(ii\right)$
similarly,  $k_2=\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_0}\mathrm{....}\left(ii\right)$
subtracting eq.$\left(ii\right)$  from eq. $\left(i\right)$, we get  
 $$\begin{gathered} 3k_1-k_2=\frac{-2hc}{{\lambda }_0}\Rightarrow k_2-3k_1=\frac{2hc}{{\lambda }_0} \\ \Rightarrow k_2-3k_1=2\varphi \\ \Rightarrow k_2>3k_1\Rightarrow k_1<\frac{k_2}{3} \end{gathered}$$