So that's how small drops evaporate!

If the radius of the drop is reduced by $dr$ then the potential energy will be reduced by $dU$ and some of the mass $dm$ will evaporate.
For this to happen, the loss in the surface energy of water should be greater than or equal to the energy needed to evaporate the mass $dm$ .
For this,
$dU \geq dm L$ .
Here, $L$ is the latent heat of vaporization of water.

Now, $dU = T ds$ , Here T is the surface tension and $ds$ is the change in the surface area.
$ds = 8 \pi r dr$ .
Similarly, $dm = \rho dV$ , Here $\rho$ is the density of water and $dV$ is change in its volume.
$dV = 4 \pi r^2 dr$ .
Hence, for evaporation,

$\begin{aligned} dU &\geq& (dm)L. \\ \Rightarrow T (ds) &\geq& (dm)L. \\ T \times(8\pi rdr) &\geq& L \times\rho\times( 4\pi r^2dr). \\ \Rightarrow r &\leq& \frac{2T}{\rho L}. \\ \end{aligned}$

Therefore, the maximum radius for the drop to evaporate is $\boxed{\dfrac{2T}{\rho L}}$