A tower T$$_1$$ of height 60 m is located exactly opposite to a tower T$$_2$$ of height 80 m on a straight road. From the top of T$$_1$$, if the angle of depression of the foot of T$$_2$$ is twice the angle of elevation of the top of T$$_2$$, then the width (in m) of the road between the feet of the towers T$$_1$$ and T$$_2$$ is:

Let us denote the foot of tower T$$_1$$ by $$D$$, its top by $$A$$, the foot of tower T$$_2$$ by $$B$$ and its top by $$C$$. The heights are $$AD = 60 \text{ m}$$ and $$BC = 80 \text{ m}$$. The two feet $$D$$ and $$B$$ lie on the straight road, so $$DB$$ is the required width of the road. We shall denote this width by $$x$$ metres.

Because both towers are perpendicular to the road, the horizontal distance from the top of T$$_1$$ to tower T$$_2$$ is also $$x$$. Thus:

From $$A$$ to $$B$$ we have a right-angled triangle with horizontal side $$AB = x$$ and vertical side $$AD = 60$$. From $$A$$ to $$C$$ we have another right-angled triangle with the same horizontal side $$AC_{\text{horizontal}} = x$$ and vertical side $$(BC - AD) = 80 - 60 = 20$$.

Let $$\theta$$ be the angle of elevation from $$A$$ to the top $$C$$. Then, according to the statement, the angle of depression from $$A$$ to the foot $$B$$ equals $$2\theta$$. Using the definition of tangent (opposite divided by adjacent) we write:

$$\tan\theta = \frac{20}{x} \qquad\text{and}\qquad \tan(2\theta) = \frac{60}{x}.$$

We now employ the double-angle formula for tangent, stated here for clarity: $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}.$$

Substituting $$\tan\theta = \dfrac{20}{x}$$ into the formula gives

$$\frac{60}{x} = \frac{2\left(\dfrac{20}{x}\right)}{1 - \left(\dfrac{20}{x}\right)^2}.$$

To remove complex fractions, set $$t = \frac{20}{x},$$ so that $$\tan\theta = t.$$ Then the two equations become

$$\tan(2\theta) = \frac{60}{x} = 3t$$ and by the double-angle formula $$\tan(2\theta) = \frac{2t}{1 - t^2}.$$

Equating the two expressions for $$\tan(2\theta)$$ we have

$$3t = \frac{2t}{1 - t^2}.$$

Because $$t \neq 0$$ (the towers are not coincident), we can divide both sides by $$t$$ to get

$$3 = \frac{2}{1 - t^2}.$$

Cross-multiplying yields

$$3(1 - t^2) = 2 \quad\Longrightarrow\quad 3 - 3t^2 = 2.$$

Rearranging, we obtain

$$3t^2 = 1 \quad\Longrightarrow\quad t^2 = \frac{1}{3}.$$

Taking the positive square root (since $$t > 0$$ for positive distances) gives

$$t = \frac{1}{\sqrt3}.$$

Remembering that $$t = \dfrac{20}{x}$$, we substitute back:

$$\frac{20}{x} = \frac{1}{\sqrt3} \quad\Longrightarrow\quad x = 20\sqrt3.$$

Therefore the width of the road between the feet of the two towers is $$20\sqrt3 \text{ metres}$$.

Hence, the correct answer is Option D.