If \(R\) is the radius of the earth and the acceleration due to gravity on the surface of earth is \(g =\pi^2 m / s ^2\)…

At altitude \(h=2R\) above Earth’s surface, the distance from Earth’s center is \(r=R+h=3R\).
Acceleration due to gravity there is
\[g' = g\left(\frac{R}{r}\right)^2 = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}.\]

For a seconds pendulum \(T=2\,\text{s}\) and
\[T=2\pi\sqrt{\frac{L}{g'}} \;\Rightarrow\; L=\frac{g' T^2}{4\pi^2}.\]
With \(g=\pi^2\,\text{m s}^{-2}\),
\[L=\frac{\left(\dfrac{g}{9}\right)\, (2)^2}{4\pi^2}=\frac{g}{9}\cdot\frac{4}{4\pi^2}=\frac{g}{9\pi^2}=\frac{\pi^2}{9\pi^2}=\frac{1}{9}\ \text{m}.\]

\[\boxed{\dfrac{1}{9}\ \text{m}}\]