Since resistance connected in arms $\mathit{CE},\mathit{ED},\mathit{CF}$
and $\mathit{FD}$ will form a balanced Wheatstone bridge,
therefore, the resistance of arm $\mathit{EF}$ becomes
ineffective. Now resistance of arm $\mathit{CED}$ or $\mathit{CFD}=2+2=4\mathrm{\Omega }$
Effective resistance of these two parallel arm $=\frac{4\times 4}{4+4}=2\mathrm{\Omega }$

Now resistance of arm $\mathit{ACDB}=2+2+2=6\mathrm{\Omega }$
in parallel with resistance arm $\mathit{AB}=2\mathrm{\Omega }$ Thus,
effective resistance between $A$ and $B$

$=\frac{6\times 2}{6+2}=\frac{3}{2}\mathrm{\Omega }$