Assume that liquid drop evaporates by decreasing in its surface energy so that its temperature remains unchanged. The surface tension of liquid drop is T, density of liquid is $\mathrm{\rho }$ and L is latent heat of vapouration of liquid. What should be the minimum radius of the drop for this to be possible 

being that,

T = surface tension

the density of the liquid: ρ

L = latent heat

We know that,

Decreased surface energy equals  heat required for vaporization

Now,

decrease in surface energy

$=\mathrm{T}\times \mathrm{\Delta A}$

$=2\mathrm{T}\times \left(4\mathrm{\pi rdr}\right)\mathrm{\_\_\_\_\_\_}\left(1\right)$

Now, the latent heat needed for vaporization

$=\mathrm{ML}$

$=\mathrm{V\rho L}$

$=4{\mathrm{\pi r}}^2\mathrm{\rho L\_\_\_\_\_\_\_}\left(2\right)$

From equation 1 and 2

$2\mathrm{T}\left(4\mathrm{\pi rdr}\right)=4{\mathrm{\pi r}}^2 \mathrm{\rho L}\Rightarrow \mathrm{r}=\frac{2\mathrm{T}}{\mathrm{\rho L}}$

Hence the correct answer is $\frac{2\mathrm{T}}{\mathrm{\rho L}}.$