Answer
$B = 5.03\times 10^{-8}~T$
Work Step by Step
We can find the magnetic field at $P_1$:
$dB = \frac{\mu_0}{4\pi}~\frac{i~ds~sin~\theta}{r^2}$
$B = \int_{-L/2}^{L/2}~\frac{\mu_0}{4\pi}~\frac{i~ds~sin~\theta}{r^2}$
$B = 2~\int_{0}^{L/2}~\frac{\mu_0}{4\pi}~\frac{i~ds~R^2}{(s^2+R^2)^{3/2}}$
$B = \frac{\mu_0~i}{2\pi~R}~\frac{s}{\sqrt{s^2+R^2}}~\Big \vert_{0}^{L/2}$
$B = \frac{\mu_0~i}{2\pi~R}~(\frac{L}{2~\sqrt{(L/2)^2+R^2}}-0)$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.0582~A)}{(2\pi)~(0.131~m)}~\frac{0.180~m}{2~\sqrt{(0.180~m/2)^2+(0.131~m)^2}}$
$B = 5.03\times 10^{-8}~T$
