A capacitor of capacitance $1\mathrm{\mu F}$  withstands a maximum voltage of 6 kV, while another capacitor of capacitance $2\mathrm{\mu F}$  withstands a maximum voltage of 4kV. If they are connected in series, the combination can withstand a maximum of 

$C_{eff}=\frac{1\times 2}{1+2}\mathrm{\mu F}=\frac{2}{3}\mathrm{\mu F}$ Capacitor 1 can store a maximum charge $1\mathrm{\mu F}\times 6\mathrm{kV}=6\mathrm{mC}$ . Capacitor 2 can store a maximum charge of $2\mathrm{\mu F}\times \text{\hspace{0.17em}4k}\mathrm{V}=\text{\hspace{0.17em}8}\mathrm{mC}$. Since they are in series the charge in each capacitor = 6mc

$\therefore {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}C}}_{eff}.V_{max}=1\mu F\times 6kV\Rightarrow V_{max}=9\text{\hspace{0.17em}kV}$