(Please pretend that this problem has a very cool name)

This was really interesting ;)

Let length l = Current flowing in the wire and R = dist. bet. the specific charge and the wire.

Distance from Wire "A" = $\frac{10}{1000}$ Distance from Wire "B" = $\frac{6}{1000}$

Provided, that the charge which experiences the magnetic field , B = $(2 \times 10^{-7}$ x $\frac{l}{R})$

Finally, B = $(2 \times 10^{-7}) \times (\frac{5}{\frac{6}{1000}})$ + $\frac{5}{\frac{10}{1000}})$ => $\boxed{4 Tesla}$

Magnetic Field experienced by a Charge:

B = ( 2 x 10^{-7} ) x (I / R), where I = Current flowing in the wire and R = the distance between the wire and the charge (in meters).

Distance from Wire A = (10 / 1000)m

Distance from Wire B = (6 / 1000)m

Using the Right Hand Thumb Rule, we realize that both the fields are going into the page. Thus, we must add the 2 magnitudes.

So, B = ( 2 x 10^{-7} ) x ( (5 / (6 / 1000)) + (5 / (10 / 1000))

which equals: 2.67E-4 Tesla into the page.