The volume of the bigger drop should be the same as the sum of the volume of two smaller drops if the radius of the bigger drop is $R\text{'}$ and volume of smaller drops is $R$ each, then, $\frac{4}{3}\pi R{\text{'}}^3=2\times \frac{4}{3}\pi R^3\Rightarrow R\text{'}={\left(2\right)}^{\frac{1}{3}}R$, so the capacitance after combination becomes, $C\text{'}=C\times 2^{\frac{1}{3}}$.