Drops of liquid of density d are floating half immersed in a liquid of density $\mathrm{\rho }$. If the surface tension of liquid is T then the radius of
the drop will be (d = density of liquid drop)

The combined effort of surface tension on the drop and buoyancy on the drop equals the weight of the drop. Thus

$\frac{4}{3}{\mathrm{\pi r}}^3\mathrm{dg}=2\mathrm{\pi rT}+\frac{1}{2}\times \frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\rho g}$

$2{\mathrm{r}}^2\mathrm{dg}=3\mathrm{T}+{\mathrm{r}}^2\mathrm{\rho g}$

$\mathrm{r}=\sqrt{\left\{\frac{3\mathrm{T}}{\mathrm{g}(2\mathrm{d}-\mathrm{\rho })}\right\}}$

Hence the correct answer is $\sqrt{\frac{3\mathrm{T}}{\mathrm{g}(2\mathrm{d}-\mathrm{\rho })}}.$