Answer
$ \Delta_{f}G^{\circ}(Ca(OH)_{2},aq)=-867.6\,kJ/mol$ which is close to the value in Appendix J, -868.07 kJ/mol.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$\implies -119.282\,kJ/mol=[\Delta_{f}G^{\circ}(C_{2}H_{2},g)+\Delta_{f}G^{\circ}(Ca(OH)_{2},aq)]-[\Delta_{f}G^{\circ}(CaC_{2},s)+2\Delta_{f}G^{\circ}(H_{2}O,l)]$
$-119.282\,kJ/mol=[209.2\,kJ/mol+\Delta_{f}G^{\circ}(Ca(OH)_{2},aq)]-[(-64.9\,kJ/mol)+2(-237.129\,kJ/mol)]$
$\implies \Delta_{f}G^{\circ}(Ca(OH)_{2},aq)=-119.282\,kJ/mol-209.2\,kJ/mol-64.9\,kJ/mol-(2\times237.129\,kJ/mol)$
$=-867.6\,kJ/mol$ which is close to the value in Appendix J, -868.07 kJ/mol.
