The length of a second pendulum at the surface of earth is $1m$. The length of second pendulum at the surface of moon, where $g$ is $\frac{1}{6}th$ that of earth's surface.

We know that the time period of a second pendulum is $2s$ .

At earth ,

                $T=2\pi \sqrt{l_e/g_e}$

                $2=2\pi \sqrt{l_e/g_e}$

or             $g_e={\pi }^2{l}_e$  ...............................eq1

At moon ,

                $T=2\pi \sqrt{l_m/g_m}$

or             $2=2\pi \sqrt{l_m/(g_e/6)}$  , given   $g_m=g_e/6$

or             $2=2\pi \sqrt{6l_m/{\pi }^2{l}_e}$   ,   putting the value of $g_m$ from  eq1

or             $l_m=l_e/6$

Now , given  $l_e=1m$

Hence ,     $l_m=1/6m$