The number of water molecules present in a drop of water weighing 0.05 mg is

1mole of ${\mathrm{H}}_2\mathrm{O} =18\mathrm{g}$

$=6.023\times {10}^{23} \mathrm{molecules}$

Mass of water drop is 0.05mg $=0.05\times {10}^{-3}\mathrm{g}$

Now 18g of ${\mathrm{H}}_2\mathrm{O} \mathrm{contains} 6.023\times {10}^{23}\mathrm{molecules}$

$\therefore 0.05\times {10}^{-3} \mathrm{g} \mathrm{of} {\mathrm{H}}_2\mathrm{O} \mathrm{would} \mathrm{contains}$

$=\frac{6.023\times {10}^{23}}{18}\times 0.05\times {10}^{-3}$

$=1.66\times {10}^{17} \mathrm{molecules}$