28 gms of $N_2$ and 6 gms of $H_2$ were heated in a closed 1 litre vessel. At equilibrium, 25.5 gms of $N{H}_3$ is present. The approximate value of $K_C$is

Consider the following reaction,$N_2+H_2\rightleftharpoons 2N{H}_3$

Initial No. of moles of $N_2$= 28/28 =1 mole

Initial moles of $H_2$ = 6/2 = 3 mole

At equilibrium,

No. of moles of $N{H}_3$= 25.5 / 17 = 1.5 mole

                         2x = 1.5

                           x = 0.75 mole

Therefore, No. of moles of $N_2$ = 1-0.75 = 0.25 mole

                  No. of moles of $H_2$ = 3(1-0.25) = 0.75 mole

 $K_C=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

$K_C=\frac{{(1.5)}^2}{(0.25){(0.75)}^3}$

 $K_C=21.33$

Hence, the correct answer is option (B) 21.33