A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound
You have to calculate the amount of substance with the mass of water.

$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$

$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$

$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$

In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.

1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.

Now you can simply multiply all values.

$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$