Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Since the charges are uniformly distributed in the ring, the net electric field at a point of $x$ that passes through its center 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
$$E_{ring,x}=\dfrac{xQ}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}}$$
Now we have two rings, as seen below, the electric field from the positively charged ring is away from its center while the electric field from the negatively charged ring is toward its center. See the figure below.
So at the middle point between the corners of the rings, the net electric field is toward the negative ring and is given by
$$E_{net}=(E_1+E_2)\hat i$$
$$E_{net}=\left[\dfrac{x|Q|}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}}+\dfrac{x|Q|}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}}\right]\hat i$$
Noting that $|Q_1|=|Q_2|=Q$, and $r_1=r_2=r=5$ cm
$$E_{net}=\left[\dfrac{2x|Q|}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}} \right]\hat i$$
Plugging the known;
$$E_{net}=\left[\dfrac{2(0.1)(20\times 10^{-9})}{4\pi (8.85 \times 10^{-12})(0.10^2+0.05^2)^{{3}/{2}}} \right]\hat i$$
$$E_{net}=(\color{red}{\bf 2.6\times 10^4}\;{\rm N/C})\hat i\tag{Rightward}$$
$$\color{blue}{\bf [b]}$$
The force exerted on $q$ is given by
$$F=qE_{net}$$
$$F=(-1.0\times 10^{-9})(2.6\times 10^4\;\hat i)$$
$$F=(\color{red}{\bf -2.6\times 10^{-5}}\;{\rm N })\hat i \tag{Leftward}$$
