Answer
See the detailed answer below.
Work Step by Step
We know that the electric field of a thin charged ring, at some point of distance $x$ from its axis, is given by
$$E_x=\dfrac{1}{4\pi\epsilon_0}\dfrac{xQ}{(x^2+R^2)^{3/2}}$$
where $R$ is the radius of the ring, see the figure below.
If the axis of the ring lies on $z$-direction, so
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{zQ}{(z^2+R^2)^{3/2}}$$
$\bullet\Rightarrow$ when $z\gt \gt R$,
the term of $(z^2+R^2)^{3/2}$ approaches $z^3$, and here is why
$z^3\bigg(1+\dfrac{R^2}{z^2}\bigg)^{3/2}\approx z^3\bigg(1+0\bigg)^{3/2}=z^3$
Thus, the electric field is then,
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{zQ}{z^3}$$
$$\boxed{E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{ Q}{z^2}}$$
This is similar to the electric field of a point charge, which is reasonable since at a long distance of $z$, the ring appears as a point charge.
$\bullet\Rightarrow$ when $z\lt \lt R$,
this means that we are measuring the electric field at a point that is so close to the center of the ring at which the net electric field approaches zero.
The term of $(z^2+R^2)^{3/2}$ approaches $R^3$, and here is why
$R^3\bigg( \dfrac{z^2}{R^2}+1\bigg)^{3/2}\approx R^3\bigg(0+1\bigg)^{3/2}=R^3$
Hence
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{zQ}{R^3}$$
where $z/R^3\approx0 $, since $z\lt \lt R$
$$\boxed{E_z=\bf 0\;\rm N/C}$$
