Suppose that a point $B$ , where net electric field is zero due to charges $8q$ and-$2$
$\overrightarrow{E}_{BO}$$=\frac{-1}{4\,\pi\,\varepsilon_{0}}. \frac{8q}{a^{2}}\hat{i}$
$\overrightarrow{E}_{BA}$$=\frac{1}{4\,\pi\,\varepsilon_{0}}. \frac{+2q}{\left(a+L\right)^{2}}\hat{i}$
According to condition $\overrightarrow{E}_{BO}$ +$\overrightarrow{E}_{BA}=0$
$\therefore \frac{-1}{4\,\pi\,\varepsilon_{0}}. \frac{8q}{a^{2}}=\frac{1}{4\,\pi\varepsilon_{0}} \frac{2q}{\left(a+L\right)^{2}}$
$\Rightarrow \frac{2}{a}=\frac{1}{a+L}$
$\Rightarrow 2a+ 2L = a$
$\therefore 2L=-a$
Thus, at distance $2L$ from oxigin, net electric field will be zero.