To solve this problem, we need to apply the principle of conservation of energy.$\\$When steam condenses and cools down, it releases heat, which is absorbed by the water.$\\$Given:$\\$• Initial mass of water $= 20$ g$\\$• Initial temperature of water $= 10^{\circ}C \\$• Final temperature of water $= 80^{\circ}C \\$• Specific heat of water $= 1$ cal g$^{-1}{}^{\circ}C^{-1} \\$• Latent heat of steam $= 540$ cal g$^{-1} \\$Let $m$ be the mass of steam condensed.$\\$The heat released by the steam when it condenses and cools down to $80^{\circ}C$ is:$\\$Heat released by condensation $= m \times 540$ cal$\\$Heat released by cooling from $100^{\circ}C$ to $80^{\circ}C = m \times 1 \times (100 - 80) \\ = 20m$ cal$\\$Total heat released by steam $= m \times 540 + 20m = 560m$ cal$\\$The heat absorbed by the water to raise its temperature from $10^{\circ}C$ to $80^{\circ}C$ is:$\\$Heat absorbed by water $= 20 \times 1 \times (80 - 10) = 1400$ cal$\\$By conservation of energy, the heat released by steam equals the heat absorbed by water:$\\ 560m = 1400 \\$Solving for $m: \\ m = \frac{1400}{560} \\ m = 2.5$ g$\\$Therefore, the total mass of water present will be:$\\$Initial mass of water $+$ mass of condensed steam $= 20 + 2.5 = 22.5$ g$\\$Thus, the mass of water present will be $22.5$ g, which corresponds to Option 4.