\(E=\frac{\sigma}{2 \epsilon_{0}} \quad(\text { given })\)

Electric field along the axis at any distance

\(x\) from the centre of the disc

\(E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right)\)

From question, \(x=R(\text { radius of disc })\)

\(\therefore \mathrm{E}^{\prime}=\) \( \frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{R}^{2}}}\right) \)

\(= \frac{\sigma}{2 \epsilon_{0}}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right) \)

\(=\frac{4}{14} \mathrm{E}\)

\(\therefore\) \(\%\) reduction in the value of electric field

\(=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%\)