Oxide ions are forming the lattice
Effective number of oxide ions $=4\left(\mathrm{c}\mathrm{c}\mathrm{p}\right)$
$\therefore$ Tetrahedral voids $=2\times 4=8$
And effective number of A atom $=\frac{1}{8}\times 8=1$
Number of octahedral voids $=4$
Effective number of B atom $=\frac{1}{2}\times 4=2$
$\therefore$ Molecular formulae $=\mathrm{A}{\mathrm{B}}_2{\mathrm{O}}_4$