Step 1: Understand how a painted cube splits.
When a big cube is painted on all six faces and then cut into \(n^3\) identical smaller cubes, the small cubes fall into four groups based on position.
Corner cubes have 3 painted faces, edge cubes (not corners) have 2 painted faces, face cubes (not edges) have 1 painted face, and inside cubes have 0 painted faces.
Step 2: Check statement I alone.
Statement I says 8 of the smaller cubes are painted on three sides.
For any cube cut into \(n^3\) pieces with \(n \geq 2\), the number of corner pieces (3 faces painted) is always 8, no matter what \(n\) is.
So this fact is true whether \(n=2\), \(n=3\), \(n=4\), or any other size. It does not tell us the value of \(n\).
Without knowing \(n\), we cannot find how many small cubes have zero painted faces, since that count is \((n-2)^3\).
So statement I alone is not enough.
Step 3: Check statement II alone.
Statement II says the number of smaller cubes is 64.
Since \(n^3 = 64\), we get \(n = 4\).
Now we can directly find the number of unpainted (interior) cubes using \((n-2)^3\):
\[ (4-2)^3 = 2^3 = 8 \]
So statement II alone gives us the full answer, 8 unpainted cubes.
Step 4: Conclusion.
Statement I alone fails because it is true for every cube size and gives no new information. Statement II alone succeeds because it fixes \(n=4\) and lets us compute the unpainted count directly.
So the question can be answered using statement II alone, but not statement I alone.
\[ \boxed{\text{Statement II alone is sufficient}} \]