Let us assume the point of contact of jockey of galvanometer be $C$, since it shows no deflection for $l=\frac{L}{2}$ thus no current will flow through the path $A C$, therefore the $6$ volt cell with internal resistance $r$ will be in parallel with $\frac{L}{2}$ arm of potentiometer wire. So, potential of both will be same.

the potential across arm $A C$ is $E_{1}=\frac{E A \rho L}{2 \rho L}=\frac{E}{2}$

As $E_{1}=6 v \Rightarrow E=12$ volts.

CASE $2: -$ when switch $S_{2}$ is closed at $l=\frac{5 L}{12}$

Similarly the resistance between $A C$ is $E_{2}=\frac{E 5 A \rho L}{12 \rho L}=\frac{5 E}{12}$

This will be equal to the emf across $6$ volt cell,

$E_{3}=6-\frac{6 r}{r+10}=E_{2}$

$\Rightarrow 5=6-\frac{6 r}{6+r}$

$\Rightarrow r=2 \Omega$