The length of a seconds pendulum at a height h = 2R from earth...

The acceleration due to gravity at height $2R$ from the surface of the Earth will be,

${g}^{'} = \dfrac{g{R}^{2}}{{\left(R + h\right)}^{2}} ⇒ {g}^{'} = \dfrac{g{R}^{2}}{{\left(R + 2R\right)}^{2}} ⇒ {g}^{'} = \dfrac{g}{9}$

Length of the second's pendulum at the surface of the Earth is $1m$.

Since, time period of a pendulum is given by,

$T = 2\pi\sqrt{\dfrac{l}{g}} ⇒ \dfrac{{T}^{'}}{T} = \sqrt{\dfrac{{l}^{'}}{l} \times \dfrac{g}{{g}^{'}}} ⇒ 1 = \sqrt{\dfrac{{l}^{'}}{l} \times \dfrac{g}{{g}^{'}}} ⇒ {l}^{'} = \dfrac{l}{9} = \dfrac{1}{9}m$