Each of the resistors shown in the network is $R$. Centre of Symmetry of the network is maintained at positive potential $V$. Current flowing in various branches will be 

Due to symmetry $V_A=V_B=V_C=V_D$  and $V_E=V_F=V_G=V_H$, So equivalent circuit can be drawn as

Equivalent resistance of the circuit is,  $R_{eq}=\frac{R}{4}+\frac{R}{8}=\frac{3R}{8}$
So, Current flowing in the equivalent circuit,  $I=\frac{V-0}{R_{eq}}=\frac{8V}{3R}$
From the equivalent circuit, 
 $I=4l_1=8l_3=l_4$

$\Rightarrow l_1=\frac{l}{4}=\frac{2V}{3R}$

And, $l_3=\frac{l}{8}=\frac{V}{3R}$

And  $l_4=l=\frac{V}{3R}$

Current in plane wires AB, BC, CD, DA is zero by symmetry i.e., $l_2=0$