Solution :
The Point c lies on the equatorial line of electric dipole formed by charge $\pm \large\frac{50}{3} $$ \times 10^{-3} \mu c$ placed at A and B respectively
Hence field intensity due to the change $q_2$
$\large\frac{1}{4 \pi \in_0}$ along BC
$\large\frac{9 \times 10^9 \times \Large\frac{50}{3} \times 10^3 \times 10^{-6}}{25 \times 10^{-4}}$
$\qquad= 6 \times 10^{4}\;NC^{-1}$
Answer : $ 6 \times 10^4 \;NC^{-1}$