Don't look at the total picture just yet, first try to understand the electric field of a single wire of length 2a. If you understand that, you can apply it to the two separate wires here and sum the result (but make sure to only count the component in the up-direction while summing!)For the electric field of a wire here is what I did, I will try to explain as much as I can:
Consider the following image:
The line (yellow) has length 2a and has a positive total charge Q, which is distributed uniformly across the wire.
Coulomb's law allows you to look at a tiny piece of the wire first and then integrate over the whole wire.
Consider the red part of the wire with length dy and charge dQ. If the linear charge density is \lambda then the charge dQ can be expressed as:
dQ = \lambda dy = \frac{Q dy}{2a}
because \lambda = \frac{Q}{2a}.
The distance from the point at dQ to point P is r, which can be expressed as:
r^2 = x^2 + y^2
Now according to Coulomb's law, the magnitude of the electric field at P due to the little wire segment at dQ is:
dE = \frac{1}{4 \pi \epsilon_0} \frac{dQ}{r^2} = \frac{Q}{4 \pi \epsilon_0} \frac{dy}{2a (x^2 + y^2)}
Note closely that this field has both x- and y-components!
If you look at the symmetry of the problem it is not hard to see that, eventually, once you take into account the whole wire, the y-components will all cancel each other out. (The top part of the wire will cancel out the y-component of the lower part, etc). Only the x-component will be present. Therefore we can now skip the y-component entirely and only concentrate on the x-component.
The x-component of dE is given in terms of the angle \theta as:
dE_x = dE \cos \theta = dE \frac{x}{\sqrt{x^2+y^2}}
Substituting dE we found before we get:
dE_x = \frac{Q}{4 \pi \epsilon_0} \frac{x \, dy}{2a (x^2 + y^2)^{3/2}}
Integrating this over the entire wire (y from -a to a) we get the following integral:
E_x = \frac{1}{4 \pi \epsilon_0} \frac{Qx}2a} \int_{-a}^a \frac{dy}{(x^2 + y^2)^{3/2}}
This is a standard integral, look it up in an integration table if you don't know it.
Calculating the integral we finally get:
E_x = \frac{1}{4 \pi \epsilon_0} \frac{Q}{x \sqrt{x^2 + a^2}}
So this is the electric field at a radial distance x from the center of a wire with length 2a and uniform charge Q.
Now what stops you from using this equation exactly to calculate the electric field at the point A in your image? Surely you can calculate the total charge Q and you know the distance x for both wires.
What seems to be bothering you is that the point A is not directly above the wire. This
does not matter! The wire is completely symmetrical about it's axis so you can rotate it about it's axis without anything changing. Just make sure to use the right distances in each calculation and you will be fine!
Try calculating the electric field at point A due to one of the 'L' wires (length 2a = 0.12m). I got an electric field of:
E = 13901.6 N/C
(Note, this is not only the up-component yet)
See if you can get the same.
You only have to sum them up afterwards, although there is one catch.
Just like I used the cosine of the angle in the above derivation, you also have to take into account the fact that the field of one of the wires in point A is not directed upward but rather a little to the side. You only need to sum the upward component of the field since due to the symmetry, all the 'to the side' components will cancel out. (This is only a tiny change in the answer though, but that's only because the angle is so small!)
