The resistors $3\mathit{\Omega }$ and $6\mathit{\Omega }$ are in parallel. If their equivalent resistance is ${\mathit{R}}_{\mathit{p}}$, then
$\frac{1}{{\mathit{R}}_{\mathit{p}}}=\frac{1}{3}+\frac{1}{6}$
$\frac{1}{{\mathit{R}}_{\mathit{p}}}=\frac{2+1}{6}=\frac{3}{6}$
$\frac{1}{{\mathit{R}}_{\mathit{p}}}=\frac{1}{2}$
or ${\mathrm{R}}_{\mathrm{p}}=2\mathit{\Omega }$
Now, ${\mathit{R}}_{\mathit{p}}$ and $5\mathit{\Omega }$ are in series,
So, ${\mathrm{R}}_{\mathrm{}\mathit{s}}=2+5$
$=7\mathit{\Omega }$
Thus the total resistance across $\mathrm{AB}=7\mathit{\Omega }$