Solution:

Let \(r\) be the radius of a small drop and \(R\) be the radius of the big drop. Volume conservation: \(8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 ⇒ R = 2r\). Capacitance of a spherical drop is \(C = 4\pi\epsilon_0 R\). So, \(C_{\text{big}} = 4\pi\epsilon_0 (2r) = 2 (4\pi\epsilon_0 r) = 2 C_{\text{small}}\).