Charged Ring Maximum Field

The strength of electric field at the given point is $E=\dfrac{Q}{4π\epsilon_0}\times \dfrac{x}{(x^2+R^2)^{\frac{3}{2}}}$ , where $Q$ and $R$ are the charge of the ring and it's radius respectively. This is maximum when $\dfrac{dE}{dx}=0\implies x=\dfrac{R}{\sqrt 2}$ . In this problem, $R=1$ cm. So the strength of the electric field is maximum when $x=\dfrac{1}{\sqrt 2}\approx \boxed {0.707}$ cm.