$3H_2+N_2\rightleftharpoons 2NH_3$$\Rightarrow \begin{cases}n_{H_2}=0,47 mol \\n_{N_2}=0,49 mol\\n_{NH_3}=0,02 mol \end{cases}\Rightarrow \begin{cases}[H_2]=0,94 M \\ [N_2]=0,98 M\\ [NH_3]=0,04 M \end{cases}$
$\Rightarrow K=\frac{(0,04)^2}{(0,94)^3.(0,98)}\approx 0,002$