Using Kirchhoff's laws on the bridge circuit with resistances 6Ω,3Ω,3Ω,3Ω,6Ω and applying a 1V test voltage:
From the loop equation:
6i1+3(2i1−i)=3(i−i1)
15i1=6i⇒i1=52i ...(1)
From the second loop:
3(i−i1)+6i1=1
3i+3i1=1
(3+56)i=1
i=215 A =Req1V
Req=521=4.2 Ω