Water at 25 °C has a density of 0.997 g/cm³…

Step 1. The liquid. $V_m = \dfrac{18.015\ \text{g/mol}}{0.997\ \text{g/cm}^3} = 18.07\ \text{cm}^3/\text{mol}$. The volume for one molecule is $\dfrac{18.07\ \text{cm}^3}{6.022 \times 10^{23}} = 3.00 \times 10^{-23}\ \text{cm}^3$. The edge is the cube root: $3.11 \times 10^{-8}\ \text{cm} = 0.311\ \text{nm}$.

Step 2. The vapour. One molecule has $\dfrac{24\,790\ \text{cm}^3}{6.022 \times 10^{23}} = 4.12 \times 10^{-20}\ \text{cm}^3$. The edge is $3.45 \times 10^{-7}\ \text{cm} = 3.45\ \text{nm}$.

Step 3. The ratio. $\dfrac{3.45}{0.311} = 11.1$.