Volume occupied by one molecule of water  

(density $=1\mathrm{g} {\mathrm{cm}}^{-3}$) is :

The wt. of $6.023\times {10}^{23}$ molecules of water is equal to g. mol. wt. of ${\mathrm{H}}_2\mathrm{O}$ i.e. $18 \mathrm{g}$

$\therefore$ Volume occupied by $6.023\times {10}^{23}$ molecules of

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{d}=1\mathrm{g}{\mathrm{cm}}^{-3}\right)=\frac{18\mathrm{g}}{1\mathrm{g}{\mathrm{cm}}^{-3}}=18{\mathrm{cm}}^3$

Hence volume occupied by 1 molecule of ${\mathrm{H}}_2\mathrm{O}$

$=\frac{18}{6.023\times {10}^{23}}\simeq 3.0\times {10}^{-23}{\mathrm{cm}}^3.$