The following concentration were obtained for the formation of $N{H}_3$ form $N_2$ and $H_2$ at equilibrium for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

$\left[N_2\right]=1.5\times {10}^{-2}M$
$\left[H_2\right]=3.0\times {10}^{-2}M$
$\left[N{H}_3\right]=1.2\times {10}^{-2}M$

Calculate equilibrium constant.

The correct option is D $356.0$
The following reaction is taking place,
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

Given:
$\left[N_2\right]=1.5\times {10}^{-2}M$
$\left[H_2\right]=3.0\times {10}^{-2}M$
$\left[N{H}_3\right]=1.2\times {10}^{-2}M$

$K_c$ can be written as:
$K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$

$K_c=\frac{(1.2\times {10}^{-2})}{(1.5\times {10}^{-2})(3.0\times {10}^{-2})}$

$K_c=355.5$