The surface tension and vapor pressure of a liquid at $25 °C$ are $8\times {10}^{-2} N{m}^{-1}$ and $2.5\times {10}^3 Pa$ respectively. The radius of the smallest spherical water droplet which can form without evaporating at $25 °C$ is

The drop will evaporate if the water pressure on liquid, is greater than vapour pressure above the surface of liquid. Let a water droplet of radius R be formed without evaporation then

Vapour pressure = Excess pressure in a drop

$P=\frac{2S}{R}\Rightarrow R=\frac{2S}{P}=\frac{2\times 8\times {10}^{-2}}{2.5\times {10}^3}=6.4\times {10}^{-5}m =64\mu m$