We know that in the schottky defect, the equal number of anion and cation are missing, and it will decrease the ${\mathrm{Z}}_{\mathrm{eff}}$ and therefore changes in density also be observed.

${\mathrm{Z}}_{\mathrm{new}} = 4-\frac{4\times 0.1}{100} = 3.9$

Now, the new density of the crystal will be

$$\begin{gathered} \mathrm{\rho } = \frac{\mathrm{Z}\times {\mathrm{M}}_{\mathrm{A}}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3} \\ \mathrm{\rho } = \frac{3.9\times 38}{6.023\times {10}^{23}\times 125\times {10}^{-30}} \\ \mathrm{\rho } = 2.1 \end{gathered}$$

Hence, Option $\mathrm{B}$ is correct