Answer
$19.3\,kJ/mol$
Work Step by Step
Recall that $\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$\implies \Delta G^{\circ}=[\Delta G_{f}^{\circ}(I_{2},g)]-[\Delta G_{f}^{\circ}(I_{2},s)]$
$=(19.3\,kJ/mol)-(0\,kJ/mol)$
$=19.3\,kJ/mol$
