On the basis of the information available from the reaction : $\dfrac{4}{3}Al + O_2 \rightarrow \dfrac{2}{3}Al_2O_3$, $\Delta G = -827$ kJ mol$^{-1}$ of $O_2$, the minimum e.m.f. required to carry out electrolysis of $Al_2O_3$ is (F = 96500 C mol$^{-1}$)

For the electrolysis (reverse reaction) $\dfrac{2}{3}Al_2O_3 \rightarrow \dfrac{4}{3}Al + O_2$, $\Delta G = +827$ kJ mol$^{-1}$.
Number of electrons transferred: $\dfrac{4}{3}$ mol of $Al^{3+}$ gains $\dfrac{4}{3} \times 3 = 4$ mol of electrons (equivalently, 2 mol $O^{2-}$ lose 4 mol electrons to form 1 mol $O_2$). So n = 4.
$\Delta G = nFE \Rightarrow E = \dfrac{\Delta G}{nF}$
$E = \dfrac{827 \times 10^3}{4 \times 96500} = \dfrac{827000}{386000} = 2.14$ V
So the minimum e.m.f. required is 2.14 V.