In a metal oxide, oxide ions are arranged in a hcp array and the metal ion occupies octahedral holes. If one-third of octahedral voids and all tetrahedral voids are empty the metal oxide formula is (M- metal)

It is provided that oxide ions are arranged in $\mathrm{HCP}$  and we know that in $\mathrm{HCP}$ there are 6 oxide ions per unit cell. 

Numbers of octahedral voids = no. of atom per unit cell. $=6$

So, out of 6 octahedral voids there exists $\frac{1}{3}$ which are empty. 

$\text{ So, }6\times \frac{1}{3}=2\text{ voids are not there }$

So, $6-2=4$ are present

Therefore, the formula of the metal oxide will be ${\mathrm{M}}_4{\mathrm{O}}_6$ or ${\mathrm{M}}_2{\mathrm{O}}_3$.