Suppose the number of oxide ions (O$^{2-}$) in the packing = 90 ∴ Number of octahedral voids = 90 As 2/3$^{\text{rd}}$ of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions present =$\frac{2}{3} \times 90 = 60$$\therefore\;\;$ Ratio of $\ \mathrm{Fe}^{3+} : \mathrm{O}^{2-} = 60 : 90 = 2 : 3$Hence, the formula of ferric oxide is $\ \mathrm{Fe}_{2}\mathrm{O}_{3}$.