In a sodium chloride type structure of A andd B,4A atoms are removed from the edges and 4Batoms are removed from corners.The formula of the new compound will be 1)AB 2)AB2 3)A3B 4)A6B7

As given AB has NaCl structure, hence one type of atoms must be arranged in FCC pattern (at all the 8 corners and all the 6 face centers) and other type of atoms must be present on all the 12 edges as well as at the center of the unit cell.

Therefore, before removing face centered atoms:

The number of 'B' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [there are 8 'B' atoms at the corners and 6 'B' atoms at the face centers]

The number of 'A' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [there are 12 'A' atoms at the edges and 1 'A' atom at the center of the unit cell]

After removing of A and B atoms :
 

The number of 'B' atoms = (4 x 1/8) + (6 x 1/2) = 0.5 + 3 = 3.5 [there are now only 4 'A' atoms at the face centers]

The number of 'A' atoms = (8 x 1/4) + (1 x 1) = 2 + 1 = 3

So the formula of compound = A₃B_3.5
In simple whole number ratio = A₆B₇