Atoms of an element 'A' occupy $\frac{2}{3}$ tetrahedral voids in the hexagonal close packed (hcp) unit cell lattice formed by the element 'B'. The formula of the compound formed by 'A' and 'B' is

The number of atoms per unit cell for Hexagonal Closest Packed Structure is 6.

The number of atoms of element B in the unit cell is 6 .

Number of tetrahedral voids is $2\times 6=12$

Two third of tetrahedral voids are occupied. Total number of atoms of element

A is : $=12\times \frac{2}{3}=8$

The ratio of the number of atoms of element A to the number of atoms of element $B$ is  $8:6\text{ or }4:3$

Hence, the formula of the compound is ${\mathrm{A}}_4{ \mathrm{B}}_3$

Hence the correct answer is (C) ${\mathrm{A}}_4{ \mathrm{B}}_3$