In which case is the number of molecules of water maximum

(a) 18 mL water
As ${\mathrm{d}}_{{\mathrm{H}}_2\mathrm{O}}=1\mathrm{g}/\mathrm{mL}\text{ So }{\mathrm{W}}_{{\mathrm{H}}_2\mathrm{O}}=18\mathrm{g}$
                                   ${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{18}{18}=1$
                                   $\mathrm{molecules} = 1\mathrm{x} {\mathrm{N}}_{\mathrm{A}}$
(b) 0.18g of water 
${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{0.18}{18}=0.01$
$\text{ (molecules) }{\mathrm{H}}_2\mathrm{O}=0.01\times {\mathrm{N}}_{\mathrm{A}}$
$\text{ (c) }{\left({\mathrm{V}}_{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}\right)}_{\mathrm{STP}}=0.00224\mathrm{L}$
${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{\mathrm{V}}{22.4}=\frac{0.00224}{22.4}=0.0001$
Molecules = 0.0001 x N_A
$$\begin{gathered} \left(\mathrm{d}\right) {\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}} = {10}^{-3} \\ \left(\mathrm{molecules}\right){\mathrm{H}}_2\mathrm{O} = {10}^{-3} \mathrm{x} {\mathrm{N}}_{\mathrm{A}} \end{gathered}$$