b
use $\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{d} 1}=\mu_{0} \mathrm{I}_{\mathrm{enclosed}}$
Net current enclosed by path $a$ is zero
Net current enclosed by path $c$ is $A$
Net current enclosed by path $d$ is $3\, \mathrm{A}$
Net current enclosed by path $b$ is $5\, \mathrm{A}$