Answer
$B_{net} = 1.67\times 10^{-6}~T$
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 find the magnetic field at $C$ due to the arc of current with radius $R_2= 7.80~cm$:
$B = \frac{\mu_0~i~\phi}{4~\pi~R_2}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.281~A)~(\pi~rad)}{(4~\pi)~(0.0780~m)}$
$B = 1.132\times 10^{-6}~T$
By the right hand rule, this magnetic field is out of the page.
We can find the magnetic field at $C$ due to the arc of current with radius $R_1 = 3.15~cm$:
$B = \frac{\mu_0~i~\phi}{4~\pi~R_1}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(0.281~A)~(\pi~rad)}{(4~\pi)~(0.0315~m)}$
$B = 2.803\times 10^{-6}~T$
By the right hand rule, this magnetic field is into the page.
We can find the net magnetic field at $C$:
$B_{net} = (2.803\times 10^{-6}~T)-(1.132\times 10^{-6}~T)$
$B_{net} = 1.67\times 10^{-6}~T$
