From the figure, $6\mathit{\Omega }$ and $3\mathit{\Omega }$ are in parallel, so,
$\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{3}{6}=\frac{1}{2}$
$\therefore {\mathrm{R}}_{\mathrm{P}}=2\mathit{\Omega }$
Also, $4\mathit{\Omega }$ and $12\mathit{\Omega }$ are in parallel, so
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}{}^{\prime }}=\frac{1}{4}+\frac{1}{12}=\frac{3+1}{12}=\frac{4}{12}=\frac{1}{3}$
$\therefore {\mathrm{R}}_{{\mathrm{P}}^{\prime }}=3\mathit{\Omega }$
Now, the given parallel combinations are in series. So we get,
${\mathrm{R}}_{\mathrm{eq}}=2+3=5\mathit{\Omega }$