CsBr crystallise in a body centred cubic lattice. The unit cell length is $436.6\mathrm{pm}$. Given that the atomic mass of $\mathrm{Cs}=133$ and that of $\mathrm{Br}=80$amu and Avogadro number being $6.02\times {10}^{23}{ \mathrm{mol}}^{-1}$, the density of $\mathrm{CsBr}$ is

We know that the formula for density of $\mathrm{CsBr}=\frac{Z\times M}{a^3\times N_0}$

Here, $Z=1$(because 1 formula unit is presence)

$\mathrm{M}=$molar mass of $\mathrm{CsBr}=133+80=213$

$a=$ edge length of unit cell $=436.6\mathrm{pm}=436.6\times {10}^{-10} \mathrm{cm}$

Hence, the Density $=\frac{1\times 213}{{\left(436.6\times {10}^{-10}\right)}^3\times 6.02\times {10}^{23}}=4.25 \mathrm{g}/{\mathrm{cm}}^3$

The correct option is A).