The length of a seconds pendulum at a height $$h = 2R$$ from earth surface will be: (Given: $$R$$ = Radius of earth and acceleration due to gravity at the surface of earth $$g = \pi^2 \text{ m s}^{-2}$$)

We need to find the length of a seconds pendulum at height $$h = 2R$$ from Earth's surface.

Find the acceleration due to gravity at height h = 2R.

$$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} = \frac{g}{\left(1 + 2\right)^2} = \frac{g}{9}$$

Use the time period formula for a seconds pendulum.

A seconds pendulum has time period $$T = 2 \text{ s}$$:

$$T = 2\pi\sqrt{\frac{l}{g'}}$$

$$4 = 4\pi^2 \times \frac{l}{g'}$$

$$l = \frac{g'}{\pi^2} = \frac{g}{9\pi^2}$$

Substitute $$g = \pi^2 \text{ m s}^{-2}$$.

$$l = \frac{\pi^2}{9\pi^2} = \frac{1}{9} \text{ m}$$

The correct answer is Option D: $$\dfrac{1}{9} \text{ m}$$.