As, given that
Mass of a paisa coin $=0.75 \mathrm{~g}$
diagonal $=17 \mathrm{~mm}$,
As we know that,
Atomic mass of aluminium $=26.9815 \mathrm{~g}$
Avogadro's number $=6.023 \times 10^{23}$
So, the number of mole in $(0.75 \mathrm{~g})=\frac{0.75}{26.9815}$
then the number of aluminium atoms in a paisa coin,
$$
\mathrm{N}=\frac{6.023 \times 10^{23}}{26.9815} \times 0.75=1.6742 \times 10^{22}
$$
Magnitude of charge on a proton or electron
$$
=1.6 \times 10^{-19} \mathrm{C}
$$
As charge number of $\mathrm{Al}$ is 13 , each atom of $\mathrm{Al}$ contains 13 protons and 13 electrons.
$\therefore \quad$ Magnitude of positive and negative charges in one paisa coin $=\mathrm{N}$ ze
$$
\begin{aligned}
&=1.6742 \times 10^{22} \times 13 \times 1.6 \times 10^{-19} \mathrm{C} \\
&=3.48 \times 10^4 \mathrm{C}
\end{aligned}
$$
This is a very large amount of charge. So, we can conclude that even a $0.75 \mathrm{~g} \mathrm{Al}$ and $\mathrm{Mg}$ contains enormous amount of positive and negative charge (these are) equal in magnitude.