The number of octahedral voids per lattice site in a lattice is (Rounded off to the nearest integer)

A cubic closed packing (CCP) unit cell comprises atoms placed at all the corners and at the centre of all the faces of the cube. The atom present at the centre of the face is shared between two adjacent unit cells and only half of each atom belongs to an individual cell. It is also referred to as face-centred cubic (FCC).
The total number of octahedral void(s) present in a cubic close packed structure is four. Besides the body centre, there is one octahedral void at the centre of each of the twelve edges. It is surrounded by six atoms, where four atoms belong to the same unit cell (i.e. two on the corners and two 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 one-fourth of each void belongs to a particular unit cell. Thus, in a cubic close packed structure, the octahedral void at the body centre of the cube will be one.
Twelve octahedral voids are located at each edge and are shared by four-unit cells; thus, the number of octahedral voids will be $12\times \frac{1}{4}=3$
So, the total number of octahedral voids in cubic closed packed is four.
We know, in CCP structure each unit cell has four atoms. Thus, the number of octahedral voids will be $\frac{4}{4}=1$