A uniform chain of mass m and length l overhangs a table with...

Mass of the hanging portion of the chain is $m' = \dfrac{m}{3}$.

The center of mass of the hanging portion is at a depth $h = \dfrac{l/3}{2} = \dfrac{l}{6}$ below the edge of the table.

Work done to pull the hanging portion back onto the table is equal to the change in potential energy of the hanging portion.

$W = m'gh = \left(\dfrac{m}{3}\right)g\left(\dfrac{l}{6}\right) = \dfrac{mgl}{18}$