If oxide ions are arranged in hcp and the aluminium ions occupy two...

In hcp arrangement of oxide ion. Number of octabedral (oh) voids for each oxide ion $=1$ Number of tetrahedral $(\mathrm{Td}$ ) voids for each oxide ion $=2$ $\mathrm{Al}^{3+}$ occupies only $2 / 3$ of octahedral voids. Thus, number of $\mathrm{Al}^{3+}$ for each oxide ion $=2 / 3$ $\Rightarrow$ Ratio of oxide ion : Aluminium ion $\Rightarrow 3: 2$ Thus, formula is $\mathrm{Al}_2 \mathrm{O}_3$.