Answer
$2.0\times 10^{22}~molecules$
Work Step by Step
We know that the mass of the drop is
$m=density\times volume$
$m=(10^6g/m^3)\frac{4}{3}\pi (0.52\times 10^{-2}m)^3$
Molecular mass of $H_2O$ is given as
$m_{H_2O}=\frac{M}{N_A}$
$m_{H_2O}=\frac{18.015g/mol}{6.022\times 10^{23}molecules/mol}=2.99\times 10^{-23}g/molecule$
Now, the number of molecules in a rain drop can be calculated as
$n=\frac{m}{m_{H_2O}}$
We plug in the known values to obtain:
$n=\frac{(10^6g/m^3)\frac{4}{3}\pi (0.52\times 10^{-2}m)^3}{2.99\times 10^{-23}g/molecule}$
$n=2.0\times 10^{22}~molecules$
