The magnetic field due to the wire along the $z$ axis is
$\begin{align*}\mathbf{B}_1 = \frac{\mu_0 I}{2 \pi x}\hat{\mathbf{y}}\end{align*}$
The magnetic field due to the wire with a positive slope in the picture is
$\begin{align*}\mathbf{B}_2 = \frac{\mu_0 I}{2 \pi (x/\sqrt{2}}\hat{\mathbf{y}}\end{align*}$
The magnetic field due to the wire with a negative slope in the picture is also
$\begin{align*}\mathbf{B}_3 = \frac{\mu_0 I}{2 \pi (x/\sqrt{2}}\hat{\mathbf{y}}\end{align*}$
So, the total magnetic field is
$\begin{align*}\mathbf{B} = \mathbf{B}_1 + \mathbf{B}_2 +\mathbf{B}_3 = \boxed{\frac{\mu_0 I}{2\pi x}\left(1 + 2 \sqrt{2}\righ...$
Therefore, answer (C) is correct.

Solution to 2008 Problem 60