Answer
$B = (-5.20\times 10^{-5}~T)~\hat{k}$
Work Step by Step
We can write the expression for the magnetic field produced by a current in a straight wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
We can find the magnitude of the net magnetic field at the origin:
$B = \frac{\mu_0~i_1}{2~\pi~R_1}+\frac{\mu_0~i_2}{2~\pi~R_2}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(6.00~A)}{(2\pi)~(0.100~m)}+\frac{(4\pi\times 10^{-7}~H/m)~(10.0~A)}{(2\pi)~(0.0500~m)}$
$B = (1.20\times 10^{-5}~T)+(4.00\times 10^{-5}~T)$
$B = 5.20\times 10^{-5}~T$
By the right hand rule, this magnetic field is directed in the -z direction.
We can express the net magnetic field in unit-vector notation:
$B = (-5.20\times 10^{-5}~T)~\hat{k}$
