Atom $\mathrm{B}$ forms HCP as anions are larger in size.

From the formula ${\mathrm{A}}_2{\mathrm{B}}_3$,

Number of $\mathrm{B}$ atoms$=3$,

Then, number of tetrahedral voids $=2\times$Tetrahedral voids$=2\times 3=6$.

As $\text{'}\mathrm{A}\text{'}$ is occupying tetrahedral voids,

 $\frac{1}{3}$ of tetrahedral voids$=\frac{1}{3}\times 6=2$

Therefore, formula of the compound$={\mathrm{A}}_2{\mathrm{B}}_3$