A small spherical droplet of density $$d$$ is floating exactly half immersed in a liquid of density $$\rho$$ and surface tension $$T$$. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

Let the radius of the spherical droplet be $$r$$. We start by writing every force that acts on the droplet while it floats exactly half-immersed in the liquid.

First, the weight of the droplet. The volume of a sphere is $$\frac{4}{3}\pi r^{3}$$, so using the definition $$\text{weight} = \text{(mass)}\,g = (\text{density}\times\text{volume})\,g$$ we have

$$W = \frac{4}{3}\pi r^{3}\,d\,g.$$

Next, the buoyant force. Archimedes’ principle states that the buoyant force equals the weight of the liquid displaced. Because the droplet is half-immersed, only half its volume is displaced:

$$V_{\text{submerged}} = \frac{1}{2}\left(\frac{4}{3}\pi r^{3}\right)=\frac{2}{3}\pi r^{3}.$$

Hence the upward buoyant force is

$$F_{b} = V_{\text{submerged}}\;\rho\,g = \frac{2}{3}\pi r^{3}\,\rho\,g.$$

Now we consider the force due to surface tension. Surface tension $$T$$ pulls all along the circular line of contact, whose circumference is $$2\pi r$$. For a droplet that meets the liquid surface at a right angle (which is the case when it is exactly half immersed) the entire tension around the rim contributes vertically upward. Therefore, the upward force supplied by surface tension is

$$F_{T} = T\;(2\pi r)=2\pi r\,T.$$

Because the droplet is in equilibrium, the sum of the upward forces equals the downward weight:

$$W = F_{b} + F_{T}.$$

Substituting the three expressions just obtained, we get

$$\frac{4}{3}\pi r^{3}d g = \frac{2}{3}\pi r^{3}\rho g + 2\pi r T.$$

We now perform careful algebra step by step. First divide the entire equation by $$\pi r$$ (this removes one power of $$r$$ from every term):

$$\frac{4}{3}r^{2} d g = \frac{2}{3}r^{2} \rho g + 2 T.$$

Shift the buoyant term to the left side:

$$\frac{4}{3}r^{2} d g - \frac{2}{3}r^{2} \rho g = 2 T.$$

Take $$\frac{2}{3}r^{2}g$$ common on the left:

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

Divide both sides by 2 to simplify:

$$\left(\frac{1}{3}r^{2}g\right)(2d - \rho)=T.$$

Finally solve for $$r^{2}$$:

$$r^{2}= \frac{3T}{(2d-\rho)g}.$$

Taking the positive square root (radius is always positive), we obtain

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

This matches Option D.

Hence, the correct answer is Option D.