Three capacitors $2\mu F,3\mu F$  and $5\mu F$ can withstand voltages of  3V, 2V and 1V respectively. Their series combination can withstand a maximum voltage equal to

Maximum charge that  $2\mu F$ capacitor can take  $=2\times 3=6\mu C.$

Maximum charge that $3\mu F$  capacitor can take  $=3\times 2=6\mu C.$

Maximum charge that  $5\mu F$ capacitor can take  $=5\times 1=5\mu C.$
In series, all capacitors will have same charge.
Maximum charge allowed will be  $5\mu C$.
Equivalent capacitance  $\frac{1}{C}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}$
Or,  $C=\frac{30}{31}\mu F$
$\therefore V_{\max }=\frac{5\mu C}{\frac{30}{31}\mu F}=\frac{31}{6}V$