The surface charge density of a thin charged disc of radius R is σ...

Electric field intensity at the centre of the disc. (\mathrm{E}=\frac{\sigma}{2 \epsilon_{0}} \text{ (given) }) Electric field along the axis at any distance $x$ from the centre of the disc (\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_{0}} \left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}-\mathrm{R}^{2}}\right)) From question, ($\mathrm{x}=\mathrm{R}$) (radius of disc) ($$\begin{aligned} &\therefore E^{\prime}=\frac{\sigma}{2 \epsilon_{0}} \left(1-\frac{R}{\sqrt{R^{2}+R^{2}}}\right)\\ &=\frac{\sigma}{2 \epsilon_{0}} \left(\frac{\sqrt{2} R-R}{\sqrt{2} R}\right)\\ &=\frac{4}{14}E \end{aligned}$$) (\therefore) \% reduction in the value of electric field $(=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14}$\%=70.7 \%)