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=4πr2,
∴ decrease in surface area is,
dA=4π(2rdr)
∴ Decrease in surface energy during time dt is,
dU=TdA=T⋅8πrdr...(1)
Decrease in volume during time dt is,
dV=4πr2dr
∴ Decrease in mass during time dt is,
dm=ρdV=4πρr2dr
∴ Heat required in vaporisation is,
dQ=Ldm=4πρr2drL...(2)
Apply conservation of energy,
decrease in surface energy=heat required in vaporisation.
⇒dU=dQ
⇒8Tπrdr=4πρr2drL
⇒2T=ρrL
⇒r=ρL2T
Hence, the minimum radius is ρL2T