\(E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d^{2}}\)
Where, \(q=\) Net charge \(=1.5 \times 10^{3} \,N / C\)
\(d =\) Distance from the centre \(=20\, cm =0.2 \,m\)
\(\varepsilon_{0}=\) Permittivity of free space and \(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,Nm ^{2} \,C ^{-2}\)
Therefore,
\(=6.67 \times 10^{9}\, C =6.67 \,n\,C\)
\(q=E\left(4 \pi \varepsilon_{0}\right) d^{2}=\frac{1.5 \times 10^{3}}{9 \times 10^{9}}\)
Therefore, the net charge on the sphere is \(6.67 \,n\,C .\)