Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
This is an integration problem, as we see in the figure below, we divided the sheet into many very thin strips. We chose a very thin strip of width $dx$ on the sheet [which has a width of $L$ and an infinite length $\infty$] and this thin strip has a charge of $dq$ and exerts an electric field of $dE$.
We know that the magnitude of the electric field exerted by a line of charge at distance $r$ is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda }{r } $$
where $\lambda=\dfrac{dQ}{L}$ and $\eta=\dfrac{dQ}{A}=\dfrac{dQ}{L dx}$ $\rightarrow \eta=\dfrac{\lambda}{dx}$, and hence, $\lambda=\eta dx$.
$$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta dx }{(r-x) } \;\hat i \tag 1$$
where $r$ is the distance between the center of the sheet and the point on $x$-direction.
Now to find the electric field by the whole of the sheet, we need to integrate from $x=-L/2$ to $x=L/2$,
$$\int_0^EdE=\int_{-L/2}^{L/2}\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta dx }{(r-x) } \;\hat i $$
$$ E=\dfrac{2\eta }{4\pi \epsilon_0}\;\hat i \int_{-L/2}^{L/2}\dfrac{dx }{(r-x) } =\dfrac{2\eta }{4\pi \epsilon_0}\;\hat i \left[ -\ln(r-x)\right]_{-L/2}^{L/2} $$
$$ E =\dfrac{-2\eta }{4\pi \epsilon_0}\;\hat i \left[ \ln\left(r-\frac{L}{2}\right)- \ln\left(r-\frac{-L}{2}\right)\right] $$
$$ E =\dfrac{-2\eta }{4\pi \epsilon_0}\cdot \ln \left[ \dfrac{ r-\frac{L}{2} }{ r+\frac{ L}{2} }\right] \;\hat i$$
where $r$ is already represents the $x$-direction. So we can replace $r$ by $x$.
$$ E =\dfrac{-2\eta }{4\pi \epsilon_0}\cdot \ln \left[ \dfrac{ 2x-L }{ 2x+L }\right] \;\hat i$$
$$ \boxed{E =\dfrac{ 2\eta }{4\pi \epsilon_0}\cdot \ln \left[ \dfrac{ 2x+L }{ 2x-L }\right] \;\hat i}$$
$$\color{blue}{\bf [b]}$$
When $x\gt\gt L$, we can use the hint given by the author which is $\ln (1+u)=u$ if $u\lt\lt 1$.
Thus when $x\gt\gt L$, $\ln \left[ \dfrac{ 1+\frac{L}{2x} }{ 1-\frac{L}{2x} }\right] $ where $L/2x\lt \lt 1$.
Hence,
$$\ln \left[ \dfrac{ 1+\frac{L}{2x} }{ 1-\frac{L}{2x} }\right] =\ln\left(1+\frac{L}{2x}\right)-\ln\left(1-\frac{L}{2x}\right)=\frac{L}{2x}-\frac{-L}{2x}\\
=\frac{L}{x}$$
Plug into the boxed formula above,
$$ E =\dfrac{ 2\eta L }{(4\pi \epsilon_0)x} \;\hat i $$
where $\eta=\lambda/L$
$$ \boxed{E =\dfrac{ 2\lambda }{(4\pi \epsilon_0)x} \;\hat i }$$
which is a reasonable result, since at a very long distance from the sheet, it appears as an infinite line of charge. It is the electric field of an infinite line of charge with linear charge density of $\lambda$.
$$\color{blue}{\bf [c]}$$
See the second figure below.
