According to circuit diagram, capacitors are connected in series combination.
So, equivalent capacitance $\left(C_{eq}\right)=\frac{3 \times 6}{3+6}$
$=\frac{18}{9}=2 \,\mu F$
We know that,
Total electrostatic energy $(U)=\frac{1}{2} C_{e q} V^{2}$
$=\frac{1}{2} \times 2 \times(-3)^{2}$
$=9\, \mu J$
