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 and both are positively charged.
So the direction of the electric field from the left ring is toward the right while the direction of the electric field from the other ring is toward the left.
So at the middle point between the corners of the rings, the net electric field 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
$$E_{net}=\left[\dfrac{2x|Q|}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}} \right]\hat i=0$$
$$E_{net}=\color{red}{\bf 0}\;{\rm N/C} $$
$$\color{blue}{\bf [b]}$$
At the center of the left ring, the net electric field from the left ring at its center is zero, so the net electric field at this point will be from the right ring which will be to the left.
$$E_{net}=( -E_2)\hat i$$
$$E_{net}=\left[\dfrac{-2x|Q|}{4\pi \epsilon_0 (x^2+r^2)^{{3}/{2}}} \right]\hat i $$
Plugging the known;
$$E_{net}=\left[\dfrac{(-0.2)(20\times 10^{-9})}{4\pi (8.85 \times 10^{-12})(0.20^2+0.05^2)^{{3}/{2}}} \right]\hat i$$
$$E_{net}=(\color{red}{\bf- 4.1\times 10^3}\;{\rm N/C})\hat i\tag{Leftward}$$
