In body centered cubic unit cell, the lattice points are corners and body centre of the cube. That is the atoms are present at all the corners and at the body centered position . Thus the effective number of atoms in a Body Centered Cubic Unit Cell is `2` (One from all the corners and one at the center of the unit cell). Since in BCC the body centered atom touches the top four and the bottom four atoms, the length of the body diagonal `(sqrt 3 a)` is equal to `4r.`
The packing fraction in this case is `=(2 xx 4/3 pi r^3)/(((4r)/sqrt 3)^3) approx 0.68`
Void fraction `(VF) approx 0.32`

In FCC unit cell, the lattice sites are corners and face centres. That is in face centered cubic unit cell. the atoms are present at the corners and the face centres of the cube .The effective number of atoms in fcc is 4 (one from all the corners, 3 from all the six face centers
since each face centered atom is shared by two cubes). Since, each face centered atom touches the four corner
atoms, the face diagonal of the cube `( sqrt 2 a)` is equal to
Packing fraction `(PF)= (4xx 4/3 pi r^3)/(((4r)/sqrt 2)^3)`

Void fraction `(VF) approx 0.26`

A hexagonal primitive unit cell comprises of `3` layers A, B and A as shown in the figure. Each atom has a co - ordination number of 12. Each corner atom is shared by six unit cells, of which three are in the same layer and three are in the upper/ lower layer. Thus, each corner atom makes `1 //6^(th)` contribution to a unit cell. There are `12` such corner atoms, so the effective contribution from corner atoms would be `1//6 * 12 = 2`. There are two atoms at the centre of top hexagonal face and the bottom hexagonal face. They would contribute half, as they are shared between two unit cells. Three atoms of B layer are completely inside the unit cell. Thus, the total number of effective atoms in Hexagonal Primitive unit cell would be

`(1/6 xx 12)+(1/2 xx 2) +3=6`

packing fraction `(PF)=(6xx 4/3 pi r^3)/(6 xx sqrt 3/4 (2r)^2 xx 4r sqrt(2/3)) approx 0.74`

Void fraction `(VF) approx 0.26`