According to Einstein’s photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

${K_{{{\max }_1}}}$ = ${{hc} \over \lambda }$ - $\phi$₀

and that in the second case is

${K_{{{\max }_2}}}$ = ${{hc} \over {{\lambda \over 2}}}$ - $\phi$₀

Given ${K_{{{\max }_2}}}$ = 3${K_{{{\max }_1}}}$

$\therefore$ ${{2hc} \over \lambda }$ - $\phi$₀ = 3$\left( {{{hc} \over \lambda } - {\phi _0}} \right)$

$\Rightarrow$ ${{2hc} \over \lambda }$ - $\phi$₀ = ${{{3hc} \over \lambda } - 3{\phi _0}}$

$\Rightarrow$ 2$\phi$₀ = ${{{hc} \over \lambda }}$

$\Rightarrow$ $\phi$₀ = ${{hc} \over {2\lambda }}$