The total number of octahedral void(s) per atom present in a cubic close packed structure is:

The correct option is A $1$
The total number of octahedral void(s) per atom present in a cubic close packed structure is $4$. Besides the body centre, there is one octahedral void at the centre of each of the $12$ edges. It is surrounded by six atoms, four belonging to the same unit cell ($2$ on the corners and $2$ on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only $\frac{1}{4}$th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is $1$.
$12$ octahedral voids located at each edge and shared between four unit cells$=12\times \frac{1}{4}=3$
Total number of octahedral voids $=4$
We know that in ccp structure, each unit cell has $4$ atoms. Thus, the number of octahedral voids is equal to $4/4=1$.