Solution:

The hanging part of the chain has weight

$W_h = \frac{m}{3} g$

, and the part on the table has normal force

$N = \frac{2m}{3} g$

. In limiting equilibrium, friction

$F_f = \mu N$

balances

$W_h$

, so

$$\mu \times \frac{2m}{3} g = \frac{m}{3} g$$

Solving,

$\mu = \frac{1}{2}$

.