To find the solubility product $(K_{sp})$ of $Ag_2CO_3(s),$ we will use the relation between $\Delta G^{\circ}$ and $K_{sp}. \\$The relationship is given by:$\\ \Delta G^{\circ} = -RT \ln K_{sp} \\$Given:$\\ \Delta G^{\circ} = +63.3 \, \mathrm{kJ/mol} = 63300 \, \mathrm{J/mol} \\ R = 8.314 \, \mathrm{J/K \cdot mol} \\ T = 25^{\circ}C = 298 \, \mathrm{K} \\$Substitute the known values into the equation:$\\ 63300 = -8.314 \times 298 \ln K_{sp} \\$Calculate $-RT: \\ -RT = -8.314 \times 298 = -2477.572 \\$Substitute back into the equation:$\\ 63300 = -2477.572 \ln K_{sp} \\$Solve for $\ln K_{sp}: \\ \ln K_{sp} = \frac{63300}{-2477.572} \\ \ln K_{sp} = -25.549 \\$To find $K_{sp},$ take the exponential of both sides:$\\ K_{sp} = e^{-25.549} \\ K_{sp} \approx 8.0 \times 10^{-12} \\$Therefore, the solubility product $K_{sp}$ of $Ag_2CO_3(s)$ is $8.0 \times 10^{-12}. \\$This corresponds to Option 2.