When ions along a body diagonal in NaCl unit cell are absent, then two Cl− ions at the corner and one Na+ ion at the body center is removed.
So, effective number of Na+=4−1=3
Volume of Na+=43πr3+×3
Effective number of Cl−=4−(2×18)=154
Volume of Cl−=43πr3−×154
Edge length of unit cell (a)=2r++2r−
Volume of unit cell =a3=(2r++2r−)3
So the packing fraction is
P.F.=Volume of effective number of cations and anionsVolume of unit cell=43πr3+×3+43πr3−×154(2r++2r−)3=43π(r3+×3+r3−×154)8(r++r−)3