At 320 K, a gas A$$_2$$ is 20% dissociated to A(g). The standard Gibbs free energy change at 320 K and 1 atm in J mol$$^{-1}$$ is approximately: (R = 8.314 J K$$^{-1}$$ mol$$^{-1}$$; ln 2 = 0.693; ln 3 = 1.098)

We first write the balanced gaseous equilibrium

$$\mathrm{A_2(g)\; \rightleftharpoons \;2\,A(g)}$$

Let us imagine starting with 1 mol of $$\mathrm{A_2}$$. At 320 K the gas is 20 % dissociated, i.e. 0.20 mol of $$\mathrm{A_2}$$ breaks up.

Moles after dissociation:

$$\begin{aligned} \text{undissociated }\mathrm{A_2}&=&1-0.20 = 0.80 \text{ mol}\\[4pt] \text{formed }A &=&2\times 0.20 = 0.40 \text{ mol} \end{aligned}$$

Hence the total number of moles present is

$$n_{\text{total}} = 0.80 + 0.40 = 1.20 \text{ mol}$$

The total pressure is given as 1 atm, so the partial pressures equal the mole-fraction times 1 atm.

$$\begin{aligned} P_{\mathrm{A_2}} &= \frac{0.80}{1.20}\times 1 = 0.6667\ \text{atm}=\frac23\ \text{atm}\\[4pt] P_{A} &= \frac{0.40}{1.20}\times 1 = 0.3333\ \text{atm}=\frac13\ \text{atm} \end{aligned}$$

The equilibrium constant in pressure units is defined as

$$K_{p}= \frac{(P_{A})^{2}}{P_{\mathrm{A_2}}}$$

Substituting the obtained pressures,

$$K_{p}= \frac{\left(\dfrac13\right)^{2}}{\dfrac23} =\frac{1/9}{2/3} =\frac{1}{9}\times\frac{3}{2} =\frac{3}{18} =\frac16 =0.1667$$

The relation between the standard Gibbs free-energy change and the equilibrium constant is

$$\Delta G^{\circ} = -\,R\,T\,\ln K_{p}$$

We have $$R = 8.314\ \text{J K}^{-1}\text{mol}^{-1},\; T = 320\ \text{K},\; \ln\!\left(\tfrac16\right)= -\ln 6.$$ Using the provided logarithms,

$$\ln 6 = \ln 2 + \ln 3 = 0.693 + 1.098 = 1.791$$

So

$$\Delta G^{\circ}= -(8.314)(320)(-1.791) = (8.314\times320)\,(1.791)$$

First compute the product inside the brackets:

$$8.314\times320 = 2660.48$$

Now multiply by 1.791:

$$\Delta G^{\circ}= 2660.48 \times 1.791 \approx 4\,765\ \text{J mol}^{-1}$$

Rounding to the nearest integer supplied in the options, this is 4 763 J mol-1.

Hence, the correct answer is Option D.