The time period of a satellite, revolving above earth's surface at a height equal to R will be (Given $$g = \pi^2$$ m s$$^{-2}$$, $$R$$ = radius of earth)

We need to find the time period of a satellite revolving at a height $$h = R$$ above the Earth's surface, where $$R$$ is the radius of the Earth and $$g = \pi^2$$ m/s$$^2$$. The satellite is at height $$h = R$$ above the Earth's surface so the orbital radius (distance from the center of the Earth) is
$$r = R + h = R + R = 2R$$.

For a satellite of mass $$m$$ in a circular orbit the gravitational force is balanced by the centripetal force:
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ which gives the orbital velocity $$v = \sqrt{\frac{GM}{r}}$$. The time period is the circumference divided by velocity:
$$T = \frac{2\pi r}{v} = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}$$. At the Earth's surface the gravitational acceleration satisfies $$g = \frac{GM}{R^2}$$, which gives $$GM = gR^2$$. Substituting this into the time period formula yields
$$T = 2\pi\sqrt{\frac{r^3}{gR^2}}$$.

Substituting $$r = 2R$$ gives
$$T = 2\pi\sqrt{\frac{(2R)^3}{gR^2}} = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$$. With $$g = \pi^2$$ we get
$$T = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\pi \cdot \frac{\sqrt{8R}}{\pi} = 2\sqrt{8R}$$. Since $$\sqrt{8} = 2\sqrt{2}$$, it follows that
$$T = 2 \cdot 2\sqrt{2R} = 4\sqrt{2R}$$ which can also be written as
$$T = \sqrt{16 \cdot 2R} = \sqrt{32R}$$.

The correct answer is Option 3: $$\sqrt{32R}$$.