A crystalline solid substance has a density of $10 \mathrm{g}/{\mathrm{cm}}^3$ and the length of the edge of the unit cell (FCC) is $2{ \mathrm{A}}^0$. How many number of atoms are present in 200 grams of the solid ?

It is provided in the question that density of $10 \mathrm{g}/{\mathrm{cm}}^3$ , the edge length of the unit cell  $2{ \mathrm{A}}^0=2\times {10}^{-8} \mathrm{cm}$, and $Z=4$

Mathematically, the formula of density can be represented using the equation:

$\mathrm{d}=\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}$

Substituting provided and known values in the equation and solving we get,

$10=\frac{4\times \mathrm{M}}{6.023\times {10}^{23}\times {\left(2\times {10}^{-8}\right)}^3}$

$\mathrm{M}=\frac{10\times 6.023\times {10}^{23}\times {\left(2\times {10}^{-8}\right)}^3}{4}=12.046$

The number of atoms present in 200g of this solid will be:$\frac{200}{12.046}\times 6.023\times {10}^{23}=1\times {10}^{25}$.

Therefore, the correct answer is C) $1\times {10}^{25}$.