Answer
$V_1=2.8V$
Work Step by Step
We want a minimum voltage drop across capacitor $C_1$. From equation $V=\frac{Q}{C}$, we can see that voltage is inversely porportional to capacitance. Thereofre the smallest capacitance, $C_2$ will have the largest voltage drop when connected in series with the battery. To minimize voltage drop across $C_1$, the equivalent capacitance of $C_{1,2}$ must be maximum, so we connect them in parallel.
$C_{eq}=\Big(\frac{1}{2.0\mu F+3.0\mu F}+\frac{1}{1.5\mu F}\Big)^{-1}=1.15\mu F$
$Q_{eq}=Q_2=C_{eq}V=(1.15\mu F)(12V)=13.8C$
$V_2=\frac{Q_2}{C_2}=\frac{13.8C}{1.5\mu F}=9.2V$
$V_{1,3}C_{1,3}=V_2C_2$
$V_{1,3}=\frac{V_2C_2}{C_{1,3}}=\frac{(9.2V)(1.5\mu F)}{5.0\mu F}=2.8V$
Because they are connected in parallel, voltage drop across $C_1$ is equal to the voltage drop across $C_3$
