In the spinel structure, oxide ions form CCP whereas ${\frac{1}{8}}^{th}$ of the tetrahedral voids are occupied by $A^{2+}$ cations and $\frac{1}{2}$ of the octahedral voids by $B^{3+}$ cations. If the oxide ions are replaced by $X^{\frac{-8}{3}}$ ions, the number of atomic vacancies per unit cell is :

Spinel structure is a CCP with 8 tetrahedral and 4 octahedral voids per unit cell.
Number of $O^{2-}$ ions (CCP) = 4
Nmber of $A^{2+}$ ions ($\frac{1}{8}$ tetrahedral voids) = 1
Number of $B^{3+}$ ions ($\frac{1}{2}$ octahedral voids) = 2
Hence, formula unit = $A{B}_2{O}_4$
As we can see, the overall charge on the unit cell is zero originally. Now when we replace $O^{2-}$ by $X^{-8/3}$, the charges change. The net charge on the cation is 8+. The net charge on the anion becomes $\frac{32}{3}-$. Since 8+ can balance 8- of the cation, the rest can be attributed to atomic vacancy.
Unbalanced charge is -8/3 which is nothing but 1 atomic vacancy.