Explanation

(c) $F_{initial} = \frac{1}{4\pi\varepsilon_{0}} \times \frac{12 \times ( - 8)}{r^{2}}$
$F_{initial} = \frac{1}{4\pi\varepsilon_{0}} \times \frac{96}{r^{2}}$
Where $r$is the distance between them. When the charges are brought in contact, then
$q_{1} = q_{2} = \frac{12 - 8}{2} = \frac{4}{2} = 2\mu$F
$\therefore$ $F_{final} = \frac{1}{4\pi\varepsilon_{0}} \times \frac{2 \times 2}{r^{2}} = \frac{4}{r^{2}} \times \frac{1}{4\pi\varepsilon_{0}}$
$\Longrightarrow$ $|F|_{final} = \frac{4}{r^{2}} \times \frac{1}{4\pi\epsilon_{0}}$
$\therefore$ $\frac{|F|_{initial}}{|F|_{final}} = \frac{96}{4} = 24$