Answer
See the detailed answer below.
Work Step by Step
We can consider the copper plate to be an infinite plane of charge.
We know that the electric field of an infinite plane of charge is given by
$$E=\dfrac{\eta}{\epsilon_0}$$
Let's assume that the plate is too thin to have extra charges on the edges. This means that the charge is uniformly distributed on both surfaces of the copper plate.
Recall that $\eta=q/A$, where $A=\pi r^2$
$$E=\dfrac{q}{\pi \epsilon_0 r^2}\tag 1$$
$$\color{blue}{\bf [a]}$$
At 0.1 mm above the center of the top surface of the plate, the net charge is there is $q=\frac{1}{2}Q$ where $Q$ is the total charge of the copper plate.
Plugging into (1) and then plugging the known,
$$E=\dfrac{\frac{1}{2}Q}{\pi \epsilon_0 r^2}=\dfrac{\frac{1}{2}(3.5\times 10^{-9})}{\pi (8.85\times 10^{-12})(0.05)^2}$$
$$E=\color{red}{\bf 2.52\times 10^4}\;\rm N/C\tag{Upward}$$
The direction of this electric field is away from the plate since the charge is a positive charge. And since the point at which we are measuring the electric field is above the upper surface, the direction of the electric field is upward.
$$\color{blue}{\bf [b]}$$
At the plate’s center of mass, there is no charge. So that the electric field must be zero.
$$E=\color{red}{\bf0}\;\rm N/C$$
$$\color{blue}{\bf [c]}$$
At 0.1 mm below the center of the bottom surface of the plate, we got the same electric field magnitude as at 0.1 mm above the center of the upper surface of the copper plate but the direction of the electric field here is downward.
$$E=\color{red}{\bf 2.52\times 10^4}\;\rm N/C\tag{Downward}$$
