$K_H$ of $N_2$ is 1×10⁵ atm. The moles of $N_2$ dissolved in 10 mol water is (given pressure of air is 5 atm and mole fraction of $N_2$ in air is 0.8)

K_H of N₂ =1×10⁵ atm

Given pressure of air = 5 atm 

mole fraction of N₂ in air = 0.8

Pressure of N₂ = Pressure of air × mole fraction of N₂ 

$=5\times 0.8=4 \mathrm{atm}$

By Henry's law

 Pressure of N₂ = K_H × mole fraction of N₂ 

mole fraction = $\frac{\mathrm{Pressure} \mathrm{of} {\mathrm{N}}_2}{{\mathrm{K}}_{\mathrm{H}}}=\frac{4}{1\times {10}^5}=4\times {10}^{-5}$

For 1 mole of water of $4\times {10}^{-5}$ N₂ moles are dissolved

For 10 mole of water of $4\times {10}^{-5}\times 10$ N₂ moles are dissolved $=4\times {10}^{-4}$

Hence, the required answer is A) $4\times {10}^{-4}$