When light of a given wavelength is incident on a metallic surface,...

According to the Einstein's equation of photoelectric effect for the first case,

$\dfrac{hc}{\lambda} - ϕ = 6eV ⋯ ( i )$

And for the second case,

$\dfrac{hc}{4\lambda} - ϕ = 0 . 6eV$

Therefore,

$\dfrac{3hc}{4\lambda} = 5 . 4eV$

$⇒ \lambda = \dfrac{3hc}{4 \times 5 . 4eV} = \dfrac{3 \times 1 . 24 \times {10}^{- 6}}{4 \times 5 . 4}$

$⇒ \lambda = 1 . 72 \times {10}^{- 7}m$

$⇒$ from equation \left(i\right)$</p><p>$\dfrac{hc}{1 . 72 \times {10}^{- 7}} \times \dfrac{1}{1 . 6 \times {10}^{- 19}} - ϕ = 6eV$</p><p>$\dfrac{2 \times {10}^{- 25}}{2 . 75 \times {10}^{- 26}} - ϕ = 6$</p><p>$⇒ ϕ = \left(7 . 27 - 6\right) ≅ 1 . 2eV