Explanation

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