Let, $\lambda =$mass per unit length of the chain and length of mass hanging through the table is $x$.

Normal force applied by the table on the chain will be, $N=\left(l-x\right)\lambda g$.

Tension developed in the chain will balance the gravitational force acting in the downward direction, $T=\lambda xg$.

Maximum limiting force possible for the part of the table, $f_{max}=\mu N=\mu \left(l-x\right)\lambda g$

For equilibrium,

$f_{max}=T$,

$$\begin{gathered} \Rightarrow \mu \left(l-x\right)\lambda g=\lambda xg \\ \Rightarrow \frac{1}{2}\times \left(6-x\right)=x \Rightarrow x=2 \mathrm{m} \end{gathered}$$