Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 374: 116

Please see the work below.

We know that $\tau =0$ $\implies (\frac{1}{2}mg)a-(mg)b=0$ This simplifies to: $a=2b$ $\implies a+b=3b$ but $3b=\frac{L}{2}$ $\implies b=\frac{L}{6}$ Now $x=a+b$ We plug in the known values to obtain: $x=\frac{L}{2}+\frac{L}{6}$ $\implies x=\frac{2}{3}L$