Let $4\,\Omega,\ 4\,\Omega$ and $4\,\Omega$ are in series. $R_1=4+4+4=12\,\Omega\;\text{ . }\;$ Now $2\,\Omega,\ 2\,\Omega$ and $2\,\Omega$ are in series. $R_2=2+2+2=6\,\Omega$ Now $R_1=12\,\Omega,\ R_2=6\,\Omega$ and $4\,\Omega$ are in parallel.$\;\frac{1}{R_3}\;=\;\frac{1}{12}+\;\frac{1}{4}+\;\frac{1}{6}$$\;=\;\frac{1+3+2}{12}=\;\frac{6}{12}=\;\frac{1}{2}$$R_3\;=2\,\Omega$Now resistance between $A$ and $B,\ 5\,\Omega,\ 2\,\Omega$ and $6\,\Omega$ are in series. Resistance between A and B, $R=5+2+6$ $=13\,\Omega$Hence, Equivalent resistance between $A$ and $B=13\,\Omega$