$\therefore$ Charge on first capacitor

$q_{1}=C_{1} V_{1}=1.0 \times 6 \times 10^{3} \mu C=6000 \mu C$

Similarly, charge on $2\, nd$ capacitor

$q_{2}=C_{2} V_{2}=2.0 \times 4 \times 10^{3} \mu C=8000 \mu C$

In series combination, charge on each capacitor must be the same. As max. Charge on $C_{1}$ is $6000 \mu C$, therefore, max charge on $C_{2}$ must also be $6000 \mu C$.

Hence maximum voltage for the combination is $V^{\prime}=V_{1}+V_{2}^{\prime}=\frac{6000}{1.0}+\frac{6000}{2.0}=9000 V=9 k V$