Dissociation of a gas $A_2$ takes place according to the following chemical reaction. At equilibrium

Question

Dissociation of a gas $A_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K .
 $ \mathrm{A}_2(\mathrm{~g}) ightleftharpoons 2 \mathrm{~A}(\mathrm{~g}) $

The standard Gibbs energy of formation of the involved substances has been provided below:
 The degree of dissociation of $\mathrm{A}_2(\mathrm{~g})$ is given by $\left(x 	imes 10^{-2}ight)^{1 / 2}$ where $x=$ $\_\_\_\_$ . (Nearest integer). 
[Given: $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.3010, \log 3=0.48$ ] 
Assume degree of dissociation is not negligible.

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See detailed solution for explanation.