A body centered cubic solid is made up of two elements A and B. Atoms of  '$\mathrm{A}$' occupy two corners of the cube. Remaining positions of the unit cell are occupied by the atoms of ' $\mathrm{B}$ '. The formula of the compound is :

It is provided that  A occupies 2 corners of the cube.

Then, the number of atoms of $\mathrm{A}$ in the cell would be $\left(\frac{1}{8}\right)\times 2=\frac{1}{4}$

So, B would occupy the body center as well as the rest 6 corner,

Then, the number of atoms of $\mathrm{B}$ in the cell is $\left(\frac{1}{8}\right)\times 6+1=\frac{7}{4}$ 

Taking the ratio between $A$ and $B$ we get,

$\frac{\mathrm{A}}{\mathrm{B}}=\frac{1/4}{7/4}=\frac{1}{7}$

Hence, the required answer is ${\mathrm{AB}}_7$.