The height in terms of radius of the earth (R), at which the acceleration due to gravity becomes $$\frac{g}{9}$$, where g is acceleration due to gravity on earth's surface, is ______.

Acceleration due to gravity at a height $$h$$ above Earth’s surface is given by
$$g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^{2}}$$

The question states that this value becomes $$\dfrac{g}{9}$$, so set
$$\frac{g}{\left(1 + \frac{h}{R}\right)^{2}} = \frac{g}{9}$$

Cancel $$g$$ from both sides:
$$\left(1 + \frac{h}{R}\right)^{2} = 9$$

Take the positive square root (height is positive):
$$1 + \frac{h}{R} = 3$$

Solve for $$h$$:
$$\frac{h}{R} = 3 - 1 = 2$$
$$h = 2R$$

The required height is $$2R$$.

Option C which is: $$2R$$