When the wavelength of radiation falling on a metal is changed from...

{KE}_\max = \dfrac{hc}{\lambda} - ϕ = \dfrac{hc}{500} - ϕ .....(i)

Now,

3{KE}_\max = \dfrac{hc}{200} - ϕ

.....(ii)

From equation (i) and (ii)

$\dfrac{\left(ii\right)}{\left(i\right)} = \dfrac{3}{1} = \dfrac{\dfrac{hc}{200} - ϕ}{\dfrac{hc}{500} - ϕ}$

Put the value of $hc = 1237 . 5$ and solving $ϕ = 0 . 61eV$