The time period of a simple pendulum is given by the formula:

$$T = 2\pi \sqrt{\frac{l}{g}}$$

where $$T$$ is the time period, $$l$$ is the length of the pendulum, and $$g$$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $$h$$ above the Earth's surface can be expressed as:

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

where $$g$$ is the acceleration due to gravity at the surface of the Earth, $$R$$ is the radius of the Earth, and $$h$$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $$g$$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $$R$$ above Earth's surface is $$T_1$$, and we're to find the time period $$T_2$$ at a height of $$2R$$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $$T_1$$ and $$T_2$$.

For the initial case at height $$R$$:

$$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$$

For the new case at height $$2R$$:

$$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$$

The time period is proportional to the square root of the inverse of $$g$$, so:

$$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$

Therefore:

$$T_1 = \frac{2}{3}T_2$$

Rearranging this equation:

$$3T_1 = 2T_2$$

This corresponds to Option A.