For the reaction $N_2{O}_{4 }\rightleftharpoons 2N{O}_2\left(g\right)$, the degree of dissociation at equilibrium is 0.2 at 1 atm. Then $K_p$ will be

Consider the reaction, $N_2{O}_{4 }\rightleftharpoons 2N{O}_2\left(g\right)$

        Initial                    1                 0

At equilibrium           (1-$\alpha$)           2$\alpha$

 $P_{N{O}_2}=\frac{2X0.2}{1+0.2}X1=\frac{0.4}{1.2}=\frac{1}{3}$

$P_{N_2{O}_4}=\frac{1-0.2}{1+0.2}X1=\frac{0.8}{1.2}=\frac{2}{3}$

 Now, At equilibrium,

     $K_P=\frac{{\left[P_{N{O}_2}\right]}^2}{{\left[P_{N_2{O}_4}\right]}^2}=\frac{1X1X3}{3X3X2}=\frac{1}{6}$

 Hence, the correct answer is option (C) 1/6.