When photons of energy 4. 0eV fall on the surface of a metal A ,...

The relation between de-Broglie wavelength and kinetic energy is

$\lambda = \dfrac{h}{2K{m}_{e}}$
$\dfrac{\lambdaA}{\lambdaB} = \dfrac{{K}_{B}}{{K}_{A}} ⇒ \dfrac{1}{2} = \sqrt{\dfrac{{T}_{A} - 1 . 5}{TA}} ⇒ {T}_{A} = 2eV$

$⇒ {K}_{B} = 2 - 1 . 5 = 0 . 5eV$

So, the work-function of $B$ is
${ϕ}_{B} = 4 . 0 - 0 . 5 = 3 . 5eV$