Answer
$E = 1.52\times 10^{-8}~N/C$
Work Step by Step
In part (a), we found that $~~\lambda = -5.19\times 10^{-14}~C/m$
We can find an expression for the magnitude of the electric field at $P$:
$dE = \frac{\vert dq \vert}{4\pi~\epsilon_0~s^2}$
$dE = \int_{a}^{L+a}\frac{\vert \lambda \vert~ds}{4\pi~\epsilon_0~s^2}$
$E = -\frac{\vert \lambda \vert}{4\pi~\epsilon_0}~(\frac{1}{s})\Big \vert_{a}^{L+a}$
$E = -\frac{\vert \lambda \vert}{4\pi~\epsilon_0}~(\frac{1}{L+a}-\frac{1}{a})$
$E = \frac{\vert \lambda \vert}{4\pi~\epsilon_0}~(\frac{1}{a} - \frac{1}{L+a})$
We can find the magnitude of the electric field at $P$:
$E = \frac{\vert \lambda \vert}{4\pi~\epsilon_0}~(\frac{1}{a} - \frac{1}{L+a})$
$E = \frac{(5.19\times 10^{-14}~C/m)}{(4\pi)~(8.854\times 10^{-12}~F/m)}~(\frac{1}{50~m} - \frac{1}{50.0815~m})$
$E = 1.52\times 10^{-8}~N/C$
