Solid AB has the NaCl structure. Atom A occupy the corners of the cubic unit cell. If all the face centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

AB is of the NaCl-type (fcc systems).
Amount of A atoms
$$\begin{gathered} =8\times \frac{1}{8}\left(\text{corner}\right)s+6\times \frac{1}{2}\left(\text{faces}\right) \\ =1+3=4/\text{unit cell} \end{gathered}$$

Amount of B atoms
$$\begin{gathered} =12\times \frac{1}{4}\left(\text{edge}\right)+1\left(\text{body centre}\right)= \\ =1+3=4/\text{unit cell} \end{gathered}$$

Amount of A atoms eliminated (face centred tom of one axis)
$$\begin{gathered} =2\left(\text{faces}\right)\times \frac{1}{2}\left(\text{per edge centre share}\right)=1/\text{unit cell} \\ Number of remaining A atoms ==4-1=3/\text{unit cell} \\ Formula = {\text{A}}_{\text{3}}{\text{B}}_{\text{4}} \end{gathered}$$

Therefore, the correct option is D