Four capacitors marked with capacitances and break down voltages are connected as shown in the figure. The…

In the upper branch, the capacitor is connected in series, so they have the same charge so, the minimum charge is,$Q=C_1{V}_1=5\times 1\times {10}^{-3 }\mathrm{C}$, this charge is same for second connected capacitor so potential for the second capacitor for this charge, $V_2=\frac{Q}{C_2}=\frac{5\times {10}^{-3}}{4\times {10}^{-6}}=1.25 \mathrm{kV}$ so the net potential difference in the upper branch,

$V_U=V_1+V_2=\left(1.25+1\right) \mathrm{kV}=2.25 \mathrm{kV}$,

in the lower branch, the net potential should be greater so the safe potential difference is, $2.25 \mathrm{kV}$