Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the charges are uniformly distributed in the disk, the net electric field at a point of $x$ that passes through its center (axis) must be in the $x$-direction.
The $y$-components and $z$-components from the electric field are zeros.
Hence, the net electric field in this situation is then given by
$$E_{disk,x}=\dfrac{\eta}{2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$
where $\eta=\dfrac{Q}{A}=\dfrac{Q}{\pi r^2}$
$$E_{disk,x}=\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$
Note that the electric field from a solid disk is the sum of the electric field from a group of rings and this formula is found by integrating the formula of rind; you can see that in your textbook.
Now we have two disks, as seen below, the electric field from the positively charged disk is away from its center while the electric field from the negatively charged disk is toward its center. See the figure below.
So at the middle point between the corners of the disks, the net electric field is toward the negative disk (toward the left) and is given by
$$E_{net}=-(E_1+E_2)\hat i$$
$$E_{net}=-\left(\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]+\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]\right)\hat i$$
Noting that $|Q_1|=|Q_2|=Q$, and $r_1=r_2=r=5$ cm
$$E_{net}=-\left(\dfrac{ \color{red}{\bf\not} 2Q}{ \color{red}{\bf\not} 2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] \right)\hat i$$
$$E_{net}=- \dfrac{ Q}{ \pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] \hat i$$
Plugging the known;
$$E_{net}=- \dfrac{ (50\times 10^{-9})}{ \pi (0.05)^2(8.85\times 10^{-12})}\left[1-\dfrac{(0.10)}{\sqrt{(0.10)^2+(0.05)^2}}\right] \hat i$$
$$E_{net}=(\color{red}{\bf- 7.6\times 10^4}\;{\rm N/C})\hat i\tag{Leftward}$$
$$\color{blue}{\bf [b]}$$
The force exerted on $q$ is given by
$$F=qE_{net}$$
$$F=(-1.0\times 10^{-9})(-3.8\times 10^4\;\hat i)$$
$$F=(\color{red}{\bf 7.6\times 10^{-5}}\;{\rm N })\hat i \tag{Rightward}$$
