\(T \sin \theta=\frac{9 \times 10^{9} \times q ^{2}}{0.04}= F\)
\(\tan \theta==\frac{0.1}{\sqrt{0.24}}=\frac{ F }{ mg }\)
\(q =\frac{2 \sqrt{10}}{3 \sqrt{\sqrt{24}}} \times 10^{-8}\)
\(0.95 \times 10^{-8}=\frac{ a }{21} \times 10^{-3}\)
\(a =20\)
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}$ )