Key idea: Since both capacitors are charged to the same potential $V$, the stored energy is $E=\frac12 CV^2$ Hence, $\frac{E_1}{E_2}=\frac{C_1}{C_2}$ So we only need the equivalent capacitances of the two modified capacitors.

Original capacitor: For each unmodified capacitor, $C=\frac{\varepsilon_0 A}{d}$

Capacitor I (Fig. I): The three dielectrics of equal thickness are arranged along the separation, so they behave like three capacitors in series.

Each slab has thickness $d/3$, area $A$, dielectric constants $K_1,K_2,K_3$.

Their capacitances are $C'_1=\frac{K_1\varepsilon_0 A}{d/3}=3K_1C$ $C'_2=3K_2C$ $C'_3=3K_3C$

Since they are in series, $\frac{1}{C_1}=\frac{1}{3K_1C}+\frac{1}{3K_2C}+\frac{1}{3K_3C}$ $\frac{1}{C_1}=\frac{1}{3C}\left(\frac1{K_1}+\frac1{K_2}+\frac1{K_3}\right)$

Therefore, $C_1=\frac{3C}{\frac1{K_1}+\frac1{K_2}+\frac1{K_3}}$

Simplifying, $C_1=\frac{3CK_1K_2K_3}{K_1K_2+K_2K_3+K_3K_1}$

Capacitor II (Fig. II): The three dielectrics of equal thickness are arranged side by side across the area, so they behave like three capacitors in parallel.

Each slab has area $A/3$, thickness $d$.

Their capacitances are $C''_1=\frac{K_1\varepsilon_0 (A/3)}{d}=\frac{K_1C}{3}$ $C''_2=\frac{K_2C}{3}$ $C''_3=\frac{K_3C}{3}$

Since they are in parallel, $C_2=C''_1+C''_2+C''_3=\frac{C}{3}(K_1+K_2+K_3)$

Energy ratio: $\frac{E_1}{E_2}=\frac{C_1}{C_2}$

Substitute $C_1$ and $C_2$: $\frac{E_1}{E_2}=\frac{\dfrac{3CK_1K_2K_3}{K_1K_2+K_2K_3+K_3K_1}}{\dfrac{C}{3}(K_1+K_2+K_3)}$

$\frac{E_1}{E_2}=\frac{3CK_1K_2K_3}{K_1K_2+K_2K_3+K_3K_1}\cdot \frac{3}{C(K_1+K_2+K_3)}$

$\frac{E_1}{E_2}=\frac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_1K_2+K_2K_3+K_3K_1)}$

Matching with options: This matches Option D.

Comparison with stored correct answer: Stored correct answer = D.

Our derived answer also = D, so they agree.