Start with the $NaCl$ structure

In the conventional unit cell of $NaCl$:

• One ion forms an FCC lattice: it contributes $8\text{ corners}\times \frac18 + 6\text{ faces}\times \frac12 = 1+3=4$
• The other ion occupies all octahedral voids: $12\text{ edge centers}\times \frac14 + 1\text{ body center}\times 1 = 3+1=4$

The question explicitly refers to:

• anions $X$, with only one being central,
• cations $M$ at face centers and corners.

So here we take:

• $M$ at corners and face centers $\Rightarrow$ FCC positions
• $X$ at edge centers and body center

Count ions initially

• Cations $M$: $8\text{ corners}\times \frac18 + 6\text{ face centers}\times \frac12 = 1+3=4$
• Anions $X$: $12\text{ edge centers}\times \frac14 + 1\text{ body center}\times 1 = 3+1=4$

Apply instruction (i): Remove all anions $(X)$ except the central one

Initially anions are 12 edge-center $X$ plus 1 body-center $X$.

After removing all except the central one:

• Remaining anions $X = 1$

Apply instruction (ii): Replace all face-centered cations $(M)$ by anions $(X)$

There are 6 face-centered positions, each contributing $\frac12$.

So new anion contribution added is: $6\times \frac12 = 3$

Therefore total anions now: $1+3=4$

Corner cations are still present at this stage.

Apply instruction (iii): Remove all corner cations $(M)$

Corner cation contribution is: $8\times \frac18 = 1$

After removing them, no cation remains from corners.

Apply instruction (iv): Replace the central anion $(X)$ with cation $(M)$

The central ion was contributing 1 anion. Replace it by one cation:

• Anions decrease from $4$ to $3$
• Cations increase from $0$ to $1$

Final count in structure $Z$

• Number of anions $=3$
• Number of cations $=1$

Hence, $\frac{\text{number of anions}}{\text{number of cations}}=\frac31=3$

Comparison with stored answer

Derived answer $=3$, which matches the stored correct answer.