Electric field intensity, \(E =2.55 \times 10^{4} \,N \,C ^{-1}\)

Density of oil, \(\rho=1.26 \,gm / cm ^{3}=1.26 \times 10^{3} \,kg / m ^{3}\)

Acceleration due to gravity, \(g=9.81 \,m\, s ^{-2}\)

Charge on an electron, \(e=1.6 \times 10^{-19} \,C\)

Radius of the oil drop \(= r\)

Force \((F)\) due to electric field \(E\) is equal to the weight of the oil drop \((W)\)

\(F = W \Rightarrow E q=m g \Rightarrow\) \(Ene\) \(=\frac{4}{3} \pi r^{3} \times \rho \times g\)

Where, \(q =\) Net charge on the oil drop \(= ne\) \(m =\) Mass of the oil drop \(=\) Volume of the drop \(\times\) Density of oil \(=\frac{4}{3} \pi r^{3} \times \rho\)

\(r=\left[\frac{3 E n e}{4 \pi \rho g}\right]^{\frac{1}{3}}=\left[\frac{3 \times 2.55 \times 10^{4} \times 12 \times 1.6 \times 10^{-19}}{4 \times 3.14 \times 1.26 \times 10^{3} \times 9.81}\right]^{\frac{1}{3}}\)

\(=\left[946.09 \times 10^{-21}\right]^{\frac{1}{3}}=9.82 \times 10^{-7} \,mm\)

Therefore, the radius of the oil drop is \(9.82 \times 10^{-4} \;mm\).