$P_{N_2}=$ 1 atmosphere
$P_{N_2}=K_H\times N_2$
By Henry's law
$xN_2=\large\frac{P_{N_2}}{K_H}=\frac{1}{8.42\times 10^{-7}}$
$x{N_2}=\large\frac{n(N_2)}{n(N_2)+n(H_2O)}$
$xN_2=\large\frac{n(N_2)}{n(H_2O)}$
$n(N_2)=xN_2.n(H_2O)$
$nN_2=xN_2.\large\frac{1000}{18}$
$\Rightarrow \large\frac{1}{8.42\times 10^{-7}}$$\times 55.5$
$\Rightarrow 6.59\times 10^7$ moles
Hence concentration of nitrogen =$6.59\times 10^{7} mol L^{-1}$