Answer
See the detailed answer below.
Work Step by Step
Since the volume charge density of the slab is constant, the electric field strength must be the same at equal distances on either side of the center of the slab.
As seen in the figure below, we chose the Gaussian surfaces to be the orange cylinders which are centered at $z=0$ and have heights of $2z_0$ and $2z$
$$\color{blue}{\bf [a]}$$
When $-z_0\lt z\lt +z_0$, see the small orange cylinder,
$$\oint \vec Ed \vec A=\int_{\rm top} \vec Ed \vec A+\int_{\rm bottom} \vec Ed \vec A+\int_{\rm sides} \vec Ed \vec A=\dfrac{Q_{in}}{\epsilon_0}$$
where the electric field is perpendicular to the area vectors at any point on the sides of our Gaussian cylinder. Hence, $\int_{\rm sides} \vec Ed \vec A=0$.
$$\oint \vec Ed \vec A=\int_{\rm top} \vec Ed \vec A+\int_{\rm bottom} \vec Ed \vec A+0=\dfrac{Q_{in}}{\epsilon_0}$$
The area of the top surface and the bottom surface of the cylinder are the same.
Also, the angle between the electric field and the top surface is zero since the electric field there is upward while the angle between the electric field and the bottom surface is zero since the electric field there is downward.
Thus,
$$\oint \vec Ed \vec A=EA+EA=\dfrac{Q_{in}}{\epsilon_0}$$
$$2EA=\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$E=\dfrac{Q_{in}}{2A\epsilon_0}$$
where $Q_{in}= \rho V=Ah\rho $ where $A$ is the area of the circular surface of the cylinder and $h$ is its height; where, in this case, $h=2z$. Thus,
$$E=\dfrac{2Az \rho }{2A\epsilon_0}$$
$$\boxed{\vec E=\dfrac{ z \rho }{ \epsilon_0}\;\hat k}$$
$$\color{blue}{\bf [b]}$$
When $ z\gt z_0$, see the small orange cylinder,
$$\oint \vec Ed \vec A=\int_{\rm top} \vec Ed \vec A+\int_{\rm bottom} \vec Ed \vec A+\int_{\rm sides} \vec Ed \vec A=\dfrac{Q_{in}}{\epsilon_0}$$
where the electric field is perpendicular to the area vectors at any point on the sides of our Gaussian cylinder. Hence, $\int_{\rm sides} \vec Ed \vec A=0$.
$$\oint \vec Ed \vec A=\int_{\rm top} \vec Ed \vec A+\int_{\rm bottom} \vec Ed \vec A+0=\dfrac{Q_{in}}{\epsilon_0}$$
The area of the top surface and the bottom surface of the cylinder are the same.
Also, the angle between the electric field and the top surface is zero since the electric field there is upward while the angle between the electric field and the bottom surface is zero since the electric field there is downward.
Thus,
$$\oint \vec Ed \vec A=EA+EA=\dfrac{Q_{in}}{\epsilon_0}$$
$$2EA=\dfrac{Q_{in}}{\epsilon_0}$$
Hence,
$$E=\dfrac{Q_{in}}{2A\epsilon_0}$$
where $Q_{in}= \rho V=Ah\rho $ where $A$ is the area of the circular surface of the cylinder and $h$ is its height; where, in this case, $h=2z$, but we need to stick with the height that contains charges, which is $2z_0$ since the rest of the cylinder height contains no charge at all. Thus, $h=2z_0$
$$E=\dfrac{2Az_0 \rho }{2A\epsilon_0}$$
$$\boxed{\vec E=\dfrac{ z_0 \rho }{ \epsilon_0}\;\hat k}$$
$$\color{blue}{\bf [c]}$$
It is obvious, from the first boxed formula above, that the electric field increases linearly until $z=z_0$.
It is obvious, from the second boxed formula above, that the electric field after that remains constant.
See the graph below.
