Explanation
The resultant capacitance is,
$$\dfrac{1}{C} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{4} = \dfrac{5}{8}$$
$$C = \dfrac{8}{5} = 1.6{\rm{\mu F}}$$
Energy stored in capacitor is,
$$ E= \dfrac{1}{2}C{V^2}$$
$$ = \dfrac{1}{2} \times 1.6 \times {10^{ - 6}} \times {15^2}$$
$$= 0.8 \times 225 \times {10^{ - 6}}{\rm{J}}$$
$$= 1800\;{\rm{erg}}$$
Since the capacitors are connected in series so charge flowing through them will be same.
Energy stored in first capacitor of capacitance $$0.3\;{\rm{\mu F}}$$ is given by
$${E_1} = \frac{{{Q^2}}}{{2{C_1}}}$$
$${E_1} = \frac{{{Q^2}}}{{2 \times 0.3}}$$
Energy stored in second capacitor of capacitance $$0.6\;{\rm{\mu F}}$$ is given by
$${E_2} = \frac{{{Q^2}}}{{2{C_2}}}$$
$${E_2} = \frac{{{Q^2}}}{{2 \times 0.6}}$$
Now ratio of energy stored by the condensers is given by
$$\frac{{{E_1}}}{{{E_2}}} = \frac{2}{1}$$
For the first block
Thickness $$={{d}_{1}}$$
Dielectric constant $$={{k}_{1}}$$
Capacitance $$={{C}_{1}}$$
For second block
Thickness $$={{d}_{2}}$$
Dielectric constant $$={{k}_{2}}$$
Capacitance $$={{C}_{2}}$$
Capacitors C1 and C2 are connected in parallel. So equivalent capacitance is :$$ \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}} $$
$$ {{C}_{eq}}=\dfrac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}........(1) $$
Capacitance in parallel plate capacitor is given as
$${{C}_{1}}=\dfrac{{{k}_{1}}{{\varepsilon }_{0}}A}{{{d}_{1}}}\,\,\And \,\,{{C}_{2}}=\dfrac{{{k}_{2}}{{\varepsilon }_{0}}A}{{{d}_{2}}}$$
So, equation (1) becomes
$${{C}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}.............(2)$$
Equivalent capacitance for the combination is
$$C=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}.........(2)$$
From equation (1) and (2)
$$ \dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}} $$
$$ k=\dfrac{{{k}_{1}}{{k}_{2}}({{d}_{1}}+{{d}_{2}})}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}} $$
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