To determine which option has the maximum number of water molecules, we need to calculate the number of molecules for each option.$\\$Option 1: $0.00224 \,$L of water vapors at 1 atm and 273 K$\\$Using the ideal gas equation:$\\ PV = nRT \\$Where $P = 1 \,$atm$, \, V = 0.00224 \,$L$, \, R = 0.0821 \,$L atm K$^{-1}$mol$^{-1}, \, T = 273 \,$K$\\ n = \frac{PV}{RT} = \frac{1 \times 0.00224}{0.0821 \times 273} \\ n \approx 0.0001 \,$mol$\\$Number of molecules$= n \times N_A = 0.0001 \times 6.022 \times 10^{23} \\ \approx 6.022 \times 10^{19} \,$molecules$\\$Option 2: $0.18 \,$g of water$\\$Molar mass of water $= 18 \,$g/mol$\\ n = \frac{0.18}{18} = 0.01 \,$mol$\\$Number of molecules$= n \times N_A = 0.01 \times 6.022 \times 10^{23} \\ \approx 6.022 \times 10^{21} \,$molecules$\\$Option 3: $18 \,$mL of water$\\$Density of water $= 1 \,$g/mL$\\$Mass of water$= 18 \,$g$\\ n = \frac{18}{18} = 1 \,$mol$\\$Number of molecules$= n \times N_A = 1 \times 6.022 \times 10^{23} \\ = 6.022 \times 10^{23} \,$molecules$\\$Option 4: $10^{-3} \,$mol of water$\\$Number of molecules$= 10^{-3} \times 6.022 \times 10^{23} \\ = 6.022 \times 10^{20} \,$molecules$\\$Comparing the number of molecules in each case, Option 3 has the maximum number of molecules.$\\$Therefore, the correct answer is Option 3.