Here charges are equal and opposite, so by columb's law, force of attraction between charges is :
$$
F=\frac{1}{4 \pi \epsilon_0}\left(\frac{q_1 q_2}{r^2}\right) \text { where, }\left(q_1=q_2=q\right)
$$
As given that :
$$
\begin{aligned}
&\mathrm{q}=\pm 34.8 \mathrm{KC}=\pm 3.48 \times 10^4 \mathrm{C} \\
&\mathrm{r}_1=1 \mathrm{~cm}=10^{-2} \mathrm{~m}, \mathrm{r}_2=100 \mathrm{~m}, \mathrm{r}_3=10^6 \mathrm{~m}
\end{aligned}
$$
and $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2$
(i) $F_1=\frac{|q|^2}{4 \pi \varepsilon_0 r_1^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^{-2}\right)^2}$
$$
=1.09 \times 10^{23} \mathrm{~N}
$$
(ii)
$$
\begin{aligned}
F_2=\frac{|q|^2}{4 \pi \varepsilon_0 \mathrm{r}_2^2} &=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{(100)^2} \\
&=1.09 \times 10^{15} \mathrm{~N}
\end{aligned}
$$
(iii)
$$
\begin{aligned}
F_3=\frac{|q|^2}{4 \pi \varepsilon_0 r_3^2} &=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^6\right)^2} \\
&=1.09 \times 10^7 \mathrm{~N}
\end{aligned}
$$
So, when $\pm$ charges in ordinary neutral matter are separated as point charges, they exert an enormous force. Hence very difficult to disturb electrical neutrality of matter.