Initially, Let $[{A_2}] = 1\,M$ and $[A] = 0\,M$

After $20\%$ dissociation , $80\%$ of $A_2$ remains.

$[{A_2}] = 1 \times \frac{{80}}{{100}} = 0.8\,M$

$20\%$ of $1\,M$ is

 $1 \times \frac{{20}}{{100}} = 0.2.\,[A] = 2 \times 0.2 = 0.4\,M$

The equilibrium constant

$K = \frac{{{{[A]}^2}}}{{[{A_2}]}};$ $K = \frac{{{{[0.4]}^2}}}{{[0.8]}} = 0.2$ 

$\Delta {G^o} =  - RT\,\ln \,K =  - 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}$

                            $ \times 320\,K \times \ln \,0.2 = 4281\,J/mol$