Maximum charge capacitance $C_1$ can withstand is $Q_1=C_1{V}_1$ while capacitance $C_2$ can withstand maximum charge of $Q_2=C_2{V}_2$

When $C_{1}and{C}_2$ are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand, $Q_{max}=C_1{V}_1\because C_1{V}_1<C_2{V}_2$ ... for given values

The resultant capacitance of the combination is $C_R=\frac{C_2{C}_1}{C_1+C_2}$

$Q_{max}=6\times 10^{-3}C$ and $C_R=\frac{2}{3}\times 10^{-6}F$

Maximum voltage that the combination of capacitance can withstand is $V_{max}=\frac{Q_{max}}{C_R}$

$V_{max}=9kV$