Answer
The net magnetic field at $P$ is out of the page.
Work Step by Step
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 express $\theta = 74.0^{\circ}$ in units of radians:
$\theta = (74.0^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 1.29~rad$
We can find the magnetic field at $P$ due to the arc of current with radius $r = 13.5~cm$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.411~A)~(1.29~rad)}{(4~\pi)~(0.135~m)}$
$B = 3.927\times 10^{-7}~T$
By the right hand rule, this magnetic field is into the page.
We can find the magnetic field at $P$ due to the arc of current with radius $r = 10.7~cm$:
$B = \frac{\mu_0~i~\phi}{4~\pi~r}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.411~A)~(1.29~rad)}{(4~\pi)~(0.107~m)}$
$B = 4.955\times 10^{-7}~T$
By the right hand rule, this magnetic field is out of the page.
Since the magnitude of the magnetic field out of the page is greater than the magnitude of the magnetic field into the page, the net magnetic field at $P$ is out of the page.
