$$
\frac{Q_{1}}{30 c m} \stackrel{Q_{2}}{6}
$$
Checking the given options one by one
(i) $Q_{1}=Q_{2}=0 \cdot 4 \mathrm{C}$
The force on $Q_{1}$ due to $Q_{2}$
$$
\begin{aligned}
F &=k_{p} \frac{Q_{1} \times Q_{2}}{30 \times 10^{-2}} \\
&=k \frac{Q_{1} Q_{2} \times 100}{30} \\
&=k \frac{0-4 \times 0.4 \times 100}{30} \\
&=\frac{k \times 0 \cdot 16 \times 100}{30}
\end{aligned}
$$
(ii) When $Q_{1}=0 \cdot 8 \mathrm{C} Q_{2}=0$
$$
F=\frac{k \times Q_{1} Q_{2}}{30 \times 10^{-2}}=0
$$
(iii) when $Q_{1}=0, Q_{2}=0-8 \mathrm{C}$
$$
F=0
$$
(iv) $Q_{1}=0 \cdot 2 \mathrm{C} \mathrm{Q}_{2}=0 \cdot 6 \mathrm{C}$
$$
\begin{aligned}
F &=\frac{k \cdot 0 \cdot 2 \times 0 \cdot 6}{30 \times 10^{-2}} \\
&=\frac{k \times 0 \cdot 12 \times 10^{2}}{30}
\end{aligned}
$$
Thus we find that, in option(a), the force on
$Q_{1}$ will be maximum.
$\therefore$ correct answer is $Q_{1}=Q_{2}=0 \cdot 4 \mathrm{C}$