Atoms of element B form hcp lattice and those of element A occupy two third of tetrahedral voids. The formula of compound is

The total no of atoms in HCP lattice, $n=6 \left(B\right)$

The number of tetrahedral voids $=2\mathrm{n}=2\times 6=12$

It is provided that:

$\mathrm{A}=\frac{2}{3}\times 2\times \mathrm{N}=\frac{2}{3}\times 12=8$

So the required formula of the given compound is ${\mathrm{A}}_8{\mathrm{B}}_6$

The reduced version of this formula will be ${\mathrm{A}}_4{\mathrm{B}}_3$