$\frac{{kq}}{{{{{(r-a)}}^{2}}}}=\frac{{kq}}{{{{{(r+a)}}^{2}}}}+\frac{{2kq}}{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}\cos \theta$
$\frac{1}{{{{{(r-a)}}^{2}}}}=\frac{1}{{{{{(r+a)}}^{2}}}}+\frac{{2a}}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{3}{2}}}}}}$
$\frac{1}{{{{{(r-a)}}^{2}}}}-\frac{1}{{{{{(r+a)}}^{2}}}}=\frac{{2a}}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{3}{2}}}}}}$
$\frac{{4ra}}{{{{{\left( {{{r}^{2}}-{{a}^{2}}} \right)}}^{2}}}}=\frac{{2a}}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{3}{2}}}}}}$
$\Rightarrow \frac{{2r}}{{{{{\left( {{{r}^{2}}-{{a}^{2}}} \right)}}^{2}}}}=\frac{1}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{3}{2}}}}}}$
$\frac{{4{{r}^{2}}}}{{{{{\left( {{{r}^{2}}-{{a}^{2}}} \right)}}^{4}}}}=\frac{1}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{3}}}}$
$\Rightarrow 4{{r}^{2}}{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}^{3}}={{\left( {{{r}^{2}}-{{a}^{2}}} \right)}^{4}}$
$4{{r}^{8}}{{\left( {1+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}^{3}}={{r}^{8}}{{\left( {1-\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}^{4}}$
$4{{\left( {1+\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}^{3}}={{\left( {1-\frac{{{{a}^{2}}}}{{{{r}^{2}}}}} \right)}^{4}}$
Exact value cannot be solved in exam for this equation to be true $\left| {\frac{\text{a}}{\text{r}}} \right|>1\Rightarrow \text{a}>\text{r}$ But point charge Q lies between charges of dipole 1 hence electric field cannot be zero.
There for it should be bonus. But by solving from mathematical software, we are getting $\text{a}/\text{r}\approx 3$