In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is:

In a hexagonal close-packed (hcp) structure formed by atoms of element A, the number of tetrahedral voids is twice the number of atoms. If there are $$n$$ atoms of A, there are $$2n$$ tetrahedral voids.

Atoms of element M occupy $$\dfrac{2}{3}$$ of the tetrahedral voids. So the number of M atoms is $$\dfrac{2}{3} \times 2n = \dfrac{4n}{3}$$.

The ratio of M to A atoms is $$\dfrac{4n/3}{n} = \dfrac{4}{3}$$, so the formula is M$$_{4/3}$$A$$_1$$ or equivalently M$$_4$$A$$_3$$.

The answer is $$\boxed{\text{M}_4\text{A}_3}$$.