Answer
$U=0.27J$
Work Step by Step
Total energy stored $U$ can be determined as:
$U=\frac{1}{2}CV^2$....................eq(1)
Since the capacitors are connected in parallel, the equivalent capacitance $C$ is
$C=C_1+C_2=2+4=6\mu F$
We then put the values in eq(1) and solve:
$U=\frac{1}{2}({6\times10^{-6}}{)(300)}^2$
$U=0.27J$
