$\begin{aligned} & \frac{\mathrm{kq}}{(\mathrm{r}-\mathrm{a})^2}=\frac{\mathrm{kq}}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{kq}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)} \cos \theta \\ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}=\frac{1}{(\mathrm{r}+\mathrm{a})^2}+\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\ & \frac{1}{(\mathrm{r}-\mathrm{a})^2}-\frac{1}{(\mathrm{r}+\mathrm{a})^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}} \\ & \frac{4 \mathrm{ra}}{\left(\mathrm{r}^2-\mathrm{a}^2\right)^2}=\frac{2 \mathrm{a}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{2 r}{\left(r^2-a^2\right)^2}=\frac{1}{\left(r^2+a^2\right)^{\frac{3}{2}}} \\ & \frac{4 r^2}{\left(r^2-a^2\right)^4}=\frac{1}{\left(r^2+a^2\right)^3} \\ & \Rightarrow 4 r^2\left(r^2+a^2\right)^3=\left(r^2-a^2\right)^4 \\ & 4 r^8\left(1+\frac{a^2}{r^2}\right)^3=r^8\left(1-\frac{a^2}{r^2}\right)^4 \\ & 4\left(1+\frac{a^2}{r^2}\right)^3=\left(1-\frac{a^2}{r^2}\right)^4\end{aligned}$

Exact value cannot be solved in exam for this equation to be true

$$ \left|\frac{\mathrm{a}}{\mathrm{r}}\right|>1 \Rightarrow \mathrm{a}>\mathrm{r} $$

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 $\mathrm{a} / \mathrm{r} \approx 3$.