Answer
$B=\dfrac{\mu_{0} I}{2 \pi r}$
Work Step by Step
Suppose $r(t)=r \cos t(t) i+ r\sin (t) j$
Consider $\int_C B \cdot dr =\int_a^b B r'(t) dt$
or, $\int_C B \cdot dr =\int_0^{2 \pi} (-B \sin (t) i+ B \cos (t) j) (-r \sin (t) i+ r\cos (t) j) dt$
This implies that $\int_C B \cdot dr = \int_0^{2 \pi} Br ( \sin^2 t+ \cos^2 t) dt= (2 \pi r) B$
Apply Ampere's Law:
$\int_C B \cdot dr = 2 \pi r B= \mu_{0} I$
Hence, we have $B=\dfrac{\mu_{0} I}{2 \pi r}$
