Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
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
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{zQ}{(z^2+R^2)^{3/2}}\tag 1$$
We know that the electric field will be maximum when $dE/dz=0$, so we need to take the derivative relative to $dz$.
$$\dfrac{dE}{dz} =\dfrac{d }{dz}\left[ \dfrac{1}{4\pi\epsilon_0}\dfrac{zQ}{(z^2+R^2)^{3/2}}\right]=0$$
where $Q/4\pi \epsilon_0=\rm constant$
$$ \dfrac{d }{dz}\left[ \dfrac{z }{(z^2+R^2)^{3/2}}\right]=0$$
$$ \dfrac{ (z^2+R^2)^{3/2}-z(2z)(\frac{3}{2})(z^2+R^2)^{1/2} }{(z^2+R^2)^{5/2}} =0$$
Thus,
$$ (z^2+R^2)^{3/2}-3z^2(z^2+R^2)^{1/2} =0$$
Divide by $(z^2+R^2)^{1/2} $;
$$ (z^2+R^2) -3z^2 =0$$
Hence,
$$ 2z^2 =R^2$$
$$z=\pm\sqrt{\dfrac{R^2}{2}}=\pm \dfrac{R}{\sqrt{2}}$$
$$\boxed{x=\pm \dfrac{R}{\sqrt{2}}}$$
$$\color{blue}{\bf [b]}$$
To find the electric field strength at this point, we need to plug $|z|=R/\sqrt2$ into (1),
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{RQ}{\sqrt2\left(\dfrac{R^2}{2}+R^2\right)^{3/2}} $$
$$E_z=\dfrac{1}{4\pi\epsilon_0}\dfrac{RQ}{\sqrt2\left(\dfrac{3R^2}{2} \right)^{3/2}}=\dfrac{1}{4\pi\epsilon_0}\dfrac{RQ}{ (3)^{3/2} (2)^{-3/2}\sqrt{2}R^3 }$$
$$E_z= \dfrac{1}{4\pi\epsilon_0}\dfrac{ Q}{ 3\sqrt{3} (2)^{-1}R^2 }$$
$$\boxed{E_z= \dfrac{1}{4\pi\epsilon_0}\dfrac{2 Q}{ 3\sqrt{3} R^2 }}$$
