In fcc arrangement of A and B atoms where A atom are at corner of the unit cell and B atoms are at face centre, one of the atoms are missing from the corner of the unit cell then find the percentage of void space in the unit cell?

now usually an fcc structure has 4 atoms as follows,

But if 1 atom is missing from 1 corner than no of atom consideration should be,

3 + 1/8 * 7 = 31/8

hence volume of 31/8 atoms

4/3 * pi * r^3 = 0.7176 a^3

hence emplty space is around 28.24 %