When radiation of wavelength A is used to illuminate a metallic...

$\dfrac{hc}{\lambda} = ϕ + eV$ ...........(1)

$\dfrac{hc}{3\lambda} = ϕ + \dfrac{eV}{4}$ ---------(2)

from (1) & (2)

$\dfrac{hc}{\lambda}\left(1 - \dfrac{1}{3}\right) = \dfrac{3}{4}ev$

$\dfrac{hc}{\lambda}\dfrac{2}{3} = \dfrac{3}{4}eV$

$eV = \dfrac{8}{9}\dfrac{hc}{\lambda}$

$\dfrac{hc}{\lambda} = ϕ + \dfrac{8}{9}\dfrac{hc}{\lambda}$

$ϕ = \dfrac{hc}{9\lambda} = \dfrac{hc}{{\lambda}_{th}}$

${\lambda}_{th} = 9\lambda$

$∴ k = 9$