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
$$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda }{r } $$
where $\lambda=\dfrac{dQ}{L}$ and $\eta=\dfrac{dQ}{A}=\dfrac{dQ}{L dx}$ $\rightarrow \eta=\dfrac{\lambda}{dx}$, and hence, $\lambda=\eta dx$
$$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta dx }{r } \tag 1$$
If we found the electric field $dE$ of an identical strip to ours but in the left direction, we can find that the $x$ and $y$ components of both strips at height $z$ cancel each other, and the only remaining components are in the $+z$-direction.
Thus, the net electric field by the two infinite thin strips is given by
$$dE_{net}=2\left[\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta dx }{r } \cos\theta\right]\;\hat k$$
where $\cos\theta=z/r $
$$dE_{net}=2\left[\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta zdx }{r^2 } \right]\;\hat k$$
from the geometry of the figure below, $r=\sqrt{x^2+z^2}$
$$dE_{net}=2\left[\dfrac{1}{4\pi \epsilon_0}\dfrac{2\eta zdx }{(x^2+z^2)} \right]\;\hat k$$
Now to find the electric field by the whole the sheet, we need to integrate from $x=0$ to $x=L/2$,
$$\int_0^EdE_{net}=\int_0^{L/2} \left[\dfrac{1}{4\pi \epsilon_0}\dfrac{4\eta zdx }{(x^2+z^2)} \right]\;\hat k$$
$$E=\dfrac{4\eta z}{(4\pi \epsilon_0)}\;\hat k\int_0^L \dfrac{dx }{(x^2+z^2)} $$
$$E=\dfrac{4\eta z}{(4\pi \epsilon_0)}\;\hat k\left[ \dfrac{1}{z}\tan^{-1}\left(\dfrac{x}{z}\right)\right]_0^{L/2} $$
$$E=\dfrac{4\eta }{(4\pi \epsilon_0)}\;\hat k\left[ \tan^{-1}\left(\dfrac{L}{2z} \right)-\tan^{-1}\left(\dfrac{0}{z} \right)\right] $$
$$\boxed{E=\dfrac{4\eta }{(4\pi \epsilon_0)} \cdot\tan^{-1}\left(\dfrac{L}{2z} \right) \;\hat k}$$
$$\boxed{E=\dfrac{ \eta }{\pi \epsilon_0} \cdot\tan^{-1}\left(\dfrac{L}{2z} \right) \;\hat k}$$
From the figure below, we can see that $\tan\theta_{net}=\dfrac{L}{2z}$
$$\color{blue}{\bf [b]}$$
When $z\lt\lt L$, then $L/2z\approx \infty$, and hence $\tan^{-1}\left(\dfrac{L}{2z} \right) =\dfrac{\pi}{2}$.
So that,
$$E=\dfrac{ \eta \pi }{2\pi \epsilon_0 } \;\hat k$$
$$\boxed{E=\dfrac{ \eta }{2 \epsilon_0 } \;\hat k}\tag 2$$
which is a reasonable result, it is the electric field of an infinite plane of charge with surface charge density of $\eta$.
And when $z\gt\gt L$, then $L/2z\approx 0$, and hence $\tan^{-1}\left(\dfrac{L}{2z} \right) =\dfrac{L}{2z} $; Recall the small angle approximation where $\tan\theta\approx \theta$ in radians of course.
So that,
$$E=\dfrac{4\eta }{(4\pi \epsilon_0)} \cdot \dfrac{L}{2z} \;\hat k$$
Recalling that $\eta=\dfrac{\lambda}{L}$
$$E=\dfrac{4\lambda }{(4\pi \epsilon_0)L} \cdot \dfrac{L}{2z} \;\hat k$$
$$\boxed{E=\dfrac{1 }{(4\pi \epsilon_0) } \cdot \dfrac{2\lambda}{ z} \;\hat k}\tag 3$$
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 $\lambda$.
$$\color{blue}{\bf [c]}$$
To draw the graph, we need to figure out some point.
When $z=0$, from (2), $E=\eta/2\epsilon_0$,
and when $z=\infty$, from (3), $E=\dfrac{1 }{(4\pi \epsilon_0) } \cdot \dfrac{2\lambda}{ z} \approx 0$.
See the second figure below.
