Right choice is (a) \(\frac{∈_1 ∈_2}{∈_1+∈_2}\)
Easiest explanation: The combination is equal to two capacitors in series.
So, C = \(\frac{\Big[∈_0 ∈_1 \left(\frac{A}{0.5d}\right)\Big]\Big[∈_0 ∈_2 \left(\frac{A}{0.5d}\right)\Big]}{∈_0 ∈_1 \frac{A}{0.5d} + ∈_0 ∈_1 \frac{A}{0.5d}}\)
Hence, C is proportional to \(\frac{∈_1 ∈_2}{∈_1+∈_2}\).