The distance between charges 5.0 x ${10}^{-11}$C and -2. 7 x ${10}^{-11}$ C is 0.2 m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

If two opposite charges are separated by a certain distance, then for its 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 a distance x from -2. 7 x ${10}^{-11}$C, then for its equilibrium $\left|{\mathrm{F}}_1\right|=\left|{\mathrm{F}}_2\right|$.

$\begin{array}{l}\Rightarrow \frac{{\mathrm{kQ}}_1\mathrm{q}}{{\left(\mathrm{x}+0.2\right)}^2}=\frac{{\mathrm{kQ}}_2\mathrm{q}}{{\mathrm{x}}^2}\Rightarrow \mathrm{x}=0.556\mathrm{m}\\ \left(\mathrm{Here}\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\mathrm{and}\begin{array}{c}{\mathrm{Q}}_1=5\times {10}^{-11}\mathrm{C}\\ {\mathrm{Q}}_2=2.7\times {10}^{-11}\mathrm{C}\end{array}\right)\end{array}$