Find the distance from the center of the Earth

The pendulum is at a height above the Earth's surface equal to the Earth's radius $R$.

Hence its distance from the Earth's center is $r = R + R = 2R$

Acceleration due to gravity at height $h=R$

We know that gravitational acceleration at distance $r$ from the center is $g' = g\left(\frac{R}{r}\right)^2$

Substituting $r=2R$, $g' = g\left(\frac{R}{2R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$

Given $g = \pi^2\, \text{m s}^{-2}$ so $g' = \frac{\pi^2}{4}\, \text{m s}^{-2}$

Time period of a simple pendulum

For small oscillations, $T = 2\pi \sqrt{\frac{l}{g'}}$

Given length $l=4\,\text{m}$

Therefore, $T = 2\pi \sqrt{\frac{4}{\pi^2/4}}$

Simplify

$\frac{4}{\pi^2/4} = 4\cdot \frac{4}{\pi^2} = \frac{16}{\pi^2}$

So, $T = 2\pi \sqrt{\frac{16}{\pi^2}}$

$T = 2\pi \cdot \frac{4}{\pi} = 8\,\text{s}$

Final answer

$\boxed{8}$

Comparison with stored correct answer

The derived answer is $8$, which matches the stored correct answer.