Four capacitors marked with capacitances and breakdown voltages are...

Resultant capacitance of series combination 1 , $$\begin{aligned}& =\frac{1}{5}+\frac{1}{4}=\frac{9}{20}\\ C_{\mathrm{eq}_1} \text{ and } =\frac{20}{9}=2.25 \mu \mathrm{F} \end{aligned}$$Resultant capacitance of series combination 2, $$\begin{aligned} \text{ and } =\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\\ C_{\mathrm{eq}_2} & =\frac{6}{5}=1 \cdot2 \mu F \end{aligned}$$So, charge on upper branch is $=\frac{20}{9} \mathrm{~V}$ and charge on lower branch is $\frac{6}{5} \mathrm{~V}.
So, smallest value is 2.25 kV.