Solution:

Nuclear radius is given by \(R \propto A^{1/3}\), so the surface area is \(S \propto R^2 \propto A^{2/3}\). Thus, \(\frac{S_2}{S_1} = \left(\frac{64}{8}\right)^{2/3} = 8^{2/3} = 4\), meaning the surface area of the nucleus \(Y^{64}\) is \(4A_0\).