$\text{AB}$ crystallizes in a body centred cubic lattice with edge length '$a$ ' equal to $387\text{pm}$. The distance between two oppositely charged ions in the lattice is:

In an ionic BCC lattice ($\text{CsCl}$ type), the oppositely charged ions touch along the body diagonal.The distance between the center of the cation and the center of the anion is half the length of the body diagonal.
$$\text{Distance}=\frac{\sqrt{3}a}{2}$$
$$\text{Distance}=\frac{1.732\times 387}{2}=\frac{670.28}{2}=335.14\text{pm}\approx 335\text{pm}$$