Answer
$4.3\,kJ/mol$
Work Step by Step
Recall that $\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$\implies \Delta G^{\circ}=[\Delta G_{f}^{\circ}(CH_{3}OH,g)]-[\Delta G_{f}^{\circ}(CH_{3}OH,l)]$
$=(-162.3\,kJ/mol)-(-166.6\,kJ/mol)$
$=4.3\,kJ/mol$
