Given,
The area of the parallel plate capacitor is
[tex]A=1m^2[/tex]The plates are separated by d = 2 mm.
The dielectric constant is
[tex]\epsilon_r=5.6[/tex]The capacitance of the capacitor is calculated by the formula.
[tex]C=\frac{\epsilon_0\epsilon_rA}{d}[/tex]Substitute the given values,
[tex]\begin{gathered} C=\frac{8.85\times10^{-12}\times5.6\times1m^2}{2mm} \\ C=\frac{49.56\times10^{-12}}{2\times10^{-3}\text{ }} \\ C=24.78\times10^{-9}\text{ F} \end{gathered}[/tex]The electrical energy stored in the plates is calculated by the formula.
[tex]U=\frac{1}{2}CV^2[/tex]The potential difference between the plates is 6000 V.
[tex]\begin{gathered} U=\frac{1}{2}\times24.78\times10^{-9}\times(6000)^2 \\ U=2.59\times10^{-9}\times36\times10^6 \\ U=446.04\times10^{-3}\text{ }J \\ U=0.45\text{ J} \end{gathered}[/tex]Thus, the correct option is 0.45 J.