Answer
$B_{net} = 2.75\times 10^{-8}~T$
Work Step by Step
Note that the straight sections of wire do not contribute to the magnetic field at $P$
We can write the expression for the magnetic field due to an arc of current:
$B = \frac{\mu_0~i~\phi}{4~\pi~R}$
We can find the magnetic field at $P$ due to the arc of current with radius $r = 4.00~m$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.200~A)~(7\pi/4~rad)}{(4~\pi)~(4.00~m)}$
$B = 2.75\times 10^{-8}~T$
By the right hand rule, this magnetic field is out of the page.
We can find the magnetic field at $P$ due to the arc of current with radius $r = 2.00~m$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.200~A)~(7\pi/4~rad)}{(4~\pi)~(2.00~m)}$
$B = 5.50\times 10^{-8}~T$
By the right hand rule, this magnetic field is into the page.
We can find the net magnetic field at $P$:
$B_{net} = (5.50\times 10^{-8}~T)-(2.75\times 10^{-8}~T)$
$B_{net} = 2.75\times 10^{-8}~T$
