you can find the voltage in your system from the law of energy conservation.
When each of the capacitors lived its own life, the sum of energies was
[LATEX]W = \frac{C_1V_1^2}{2}+\frac{C_2V_2^2}{2}[/LATEX]
After the connection, we have
[LATEX]W = \frac{C_sV_s^2}{2}[/LATEX]
Thus
[LATEX]C_1V_1^2 + C_2V_2^2 = C_sV_s^2[/LATEX]
And
[LATEX]V_s=\sqrt{\frac{C_1V_1^2 + C_2V_2^2}{C_s}}[/LATEX]
[LATEX]V_s = 9 V[/LATEX]
Therefore, the charge
[LATEX]Q_s = C_sV_s[/LATEX]
[LATEX]Q_s = 18 \mu C[/LATEX]
For the series connection
[LATEX]Q_s = Q_{1c} = Q_{2c}[/LATEX]
[LATEX]V_s = V_{1c} + V_{2c}[/LATEX]
where "c" in indices means "connected"
Your can finally check the solution, using
[LATEX]V = Q /C [/LATEX]
[LATEX]V_s = 9V, V_{1c} = 6V, V_{2c} = 3V, 9 = 6+3 (!) [/LATEX]
