Answer
Requires work to be done.
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(Ti,s)+\Delta_{f}G^{\circ}(O_{2},g)]-[\Delta_{f}G^{\circ}(TiO_{2},s)]$
$=[(0\,kJ/mol)+(0\,kJ/mol)]-(-884.5\,kJ/mol)$
$=884.5\,kJ/mol$
As $\Delta _{r}G^{\circ}$ is positive, work needs to be done for this reaction to occur.
At least 884.5 kJ/mol work is required.
