It's still zero!

Let $CR$ denote complete ring, $CO$ represents cutout and $RR$ denotes remaining ring. Then by principle of superposition:- $E_{RR}+E_{CO}=E_{CR}=0$ Since Electric field due to complete ring at centre is $0$ . $E_{RR}=-E_{CO}$ Now since cut out is very small it can be treated as a point charge on which charge is given by $\left(\dfrac{q}{2\pi R}\right)\times x$ . $\therefore E_{RR}=-\dfrac{\frac{qx}{2\pi R}}{4\pi \varepsilon_0 R^2}=-\dfrac{qx}{8\pi^2\varepsilon_0 R^3}$ which is the required field at centre due to remaining ring and minus sign denotes that the direction of field due to remaining ring is radially outward towards the cutout. Hence $\large 8+2+3=\boxed{13}$ .