Two identical parallel plate capacitors, of capacitance C each, have plates of area A separated by a distance d. The space between the plates of two capacitors is filled with dielectrics of equal thickness and dielectric constants $K_1=1, K_2=2$ and $K_3=3$ as shown. If these two modified capacitors are charged by same potential, the ratio of energy stored $E_1$, (capacitor I) to $E_2$ (capacitor II) is:

$\frac{E_1}{E_2}=\frac{11}{9}$

The correct answer is Option (1) → $\frac{E_1}{E_2}=\frac{11}{9}$