A cubic solid A⁺B^– has the B^– ions arranged as below. If the A⁺ ions occupy all the edge centers, the formula of solid is [O = B^–] 

According to the given figure B^- ions are at corners and body center

No. of B^- ions =$8\times \frac{1}{8}(\mathrm{from} \mathrm{corners})+1(\mathrm{body} \mathrm{center})=2$

If the A⁺ ions occupy all the edge centers, then

No. of A⁺ ions = $12\times \frac{1}{4}=3$

Hence the formula is ${\mathrm{A}}_3{\mathrm{B}}_2$