Answer
$\int~B\cdot ds = 2.83\times 10^{-8}~T\cdot m$
Work Step by Step
We can examine the diagram carefully to see that the following currents are enclosed in the Amperian loop: 1, 3, 6, 7
By the right hand rule for the direction of integration, current that is out of the page is positive.
We can find $\int~B\cdot ds$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$\int~B\cdot ds = (\mu_0)~(i_1+i_3-i_6+i_7)$
$\int~B\cdot ds = (4\pi\times 10^{-7}~H/m)~(4.50~mA+13.5~mA-27.0~mA+31.5~mA)$
$\int~B\cdot ds = (4\pi\times 10^{-7}~H/m)~(22.5~mA)$
$\int~B\cdot ds = 2.83\times 10^{-8}~T\cdot m$
