Atom A occupies all corner and face center of unit cell and atom B in all octahedral void. As all atoms from face center from one axis is removed, hence number of atom A in unit cell remains 3 and B as 4.
$\mathrm{A}=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
$\mathrm{A}=4-2\times \frac{1}{2}=3$ after removing atoms
$\mathrm{B}=12\times \frac{1}{4}+1\times 1=4$
Formula of compound is ${\mathrm{A}}_3{\mathrm{B}}_4$