That's right, so I can do:
$$E = - \frac {πKλ} {2R} $$
$$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$
$$ L = 2 * π * R $$
Which should give me: R = 1.18962 L = 7.4746 Lambda: 8.41483E-10
Now I can pass to the next questions. Thank you :) !
Electric field for the semi-circle
$$E = - \frac {πKλ} {2R} $$
In this case E is equals to 10 N/C
Electric field for the straighten wire
$$E = 2Kλ * ( 1 - \frac {2y} {\sqrt{4y^2 + L^2}})$$
In this case E is equals to 8 N/C
What I'm searching is R, λ, and the length of the wire, so I think...