Explanation

(b) Given, $C = 2\mu F,\ C_{2} = 4\mu F,\ and\ V = 10$volt
Capacitors are connected in series
$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
$\therefore\ C = \ \frac{4 \times 2}{4 + 2} = \frac{4}{3}$
The charge of combination
$q = CV = \frac{4}{3} \times 10 = \frac{40}{3}$
The energy of $2\mu F$ capacitor
$E = \frac{1}{2} \times \frac{q^{2}}{C_{1}} = \frac{1}{2} \times \frac{1600}{9 \times 2} = \frac{400}{9}$
The energy of $4\mu F$ capacitor
$E_{2\ } = \frac{1}{2} \times \frac{q^{2}}{C_{2}} = \frac{1}{2} \times \frac{1600}{9 \times 4} = \frac{200}{9}$
The ratio of energies is
$\frac{E_{1}}{E_{2}} = \frac{\frac{400}{9}}{\frac{200}{9}} = \frac{2}{1}$