If R is the radius of the earth and the acceleration due to gravity...

Acceleration due to gravity at the given height, $g ' = \dfrac{GM}{{\left(3R\right)}^{2}} = \dfrac{1}{9}\dfrac{GM}{{\left(R\right)}^{2}} = \dfrac{1}{9}g$, where $g$ is acceleration due to gravity at the surface of the Earth.

Now, time period is given by, $T = 2\pi\sqrt{\dfrac{l}{g '}}$

Also the time period of second pendulum is $2$ sec.

Therefore, $T = 2sec$

Then, $2 = 2\pi\sqrt{\dfrac{l}{g} \times 9}$

$⇒ l = \dfrac{1}{9}m$