Answer
$B_{net} = 1.03\times 10^{-7}~T$
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.
We can find the net magnetic field at $P$:
$B_{net} = (4.955\times 10^{-7}~T)-(3.927\times 10^{-7}~T)$
$B_{net} = 1.03\times 10^{-7}~T$
