Explanation

(c) If two opposite charges are separated by a certain distance, then for it’s equilibrium a third charge should be kept outside and near the charge which is smaller in magnitude.
Here, suppose third charge $q$ is placed at distance $x$ from $- 2.7 \times 10^{- 11}C$ then for it’s equilibrium $\left| F_{1} \right| = |F_{2}|$
$\Rightarrow \frac{kQ_{1}q}{(x + 0.2)^{2}} = \frac{kQ_{2}q}{x^{2}} \Rightarrow x = 0.556\ m$
$\left\lbrack Here\ k = \frac{1}{4\pi\varepsilon_{0}}and\ Q_{1} = 5 \times 10^{- 11}C,\ Q_{2} = - 2.7 \times 10^{- 11}C \right\rbrack$