Steady State Analysis

Charge stored on 2μF capacitor = 22μC & on 1μF capacitor =15μC
Ratio of energy stored=242:225

The current through the 2 capacitors is 0A. Hence the 2 capacitors can be removed.

Equivalent e.m.f = 25-10 = 15 V

i=\frac{15}{4+6+5}=1A

\rightarrow V_Y - V_X = (10+4\times i)V-5V = 9V

\therefore q_2 = 2\times 9 = \underline{18\mu C}

Similarly,

\therefore q_1 = 1\times 15 = 15\mu C

Ratio of energy stored =

\frac{q_2^2}{q_1^2}\frac{C_1}{C_2}=\underline{18:25}