Assume that a drop of a liquid evaporates by a decrease in its...

Let the radius of the drop at time $t = t$ be $r$ and at an instant $t = t + dt$ be $r - dr$

As surface area is given by, $A = 4\pi{r}^{2}$,

$∴$ decrease in surface area is,

$dA = 4\pi\left(2rdr\right)$

$∴$ Decrease in surface energy during time $dt$ is,

$dU = TdA = T \cdot 8\pi rdr . . . \left(1\right)$

Decrease in volume during time $dt$ is,

$dV = 4\pi{r}^{2}dr$

$∴$ Decrease in mass during time $dt$ is,

$dm = ρdV = 4\piρ{r}^{2}dr$

$∴$ Heat required in vaporisation is,

$dQ = Ldm = 4\piρ{r}^{2}drL . . . \left(2\right)$

Apply conservation of energy,

decrease in surface energy=heat required in vaporisation.

$⇒ dU = dQ$

$⇒ 8T\pi rdr = 4\piρ{r}^{2}drL$

$⇒ 2T = ρrL$

$⇒ r = \dfrac{2T}{ρL}$

Hence, the minimum radius is $\dfrac{2T}{ρL}$