Answer
$\int~B\cdot ds = 4.5\times 10^{-6}~T\cdot m$
Work Step by Step
We can calculate the area of the triangle enclosed by the three line segments:
$A = \frac{1}{2}(4d)(3d)$
$A = 6d^2$
$A = (6)(0.20~m)^2$
$A = 0.24~m^2$
We can find the enclosed current:
$i_{enc} = (0.24~m^2)(15~A/m^2) = 3.6~A$
We can find $\int~B\cdot ds$:
$\int~B\cdot ds = \mu_0~i_{enc}$
$\int~B\cdot ds = (4\pi\times 10^{-7}~H/m)(3.6~A)$
$\int~B\cdot ds = 4.5\times 10^{-6}~T\cdot m$
