According to question
X− are present in Octahedral void
M+ are present in FCC lattice point
Number of cations and anions left after each sequence are
M+X−(i)44−3=1(ii)4−6×121+6×12(iii)1−13+1=4(iv)0+14−1=3
So, total number of X− in a unit cell = 3
total number of M+ in a unit cell = 1
Hence anioncation=31=3