Multiples of 9 that are perfect cubes: we need \(n^3\) divisible by 9, i.e., \(n^3\) divisible by \(3^2\), which means \(n\) divisible by 3.
Values: \(n = 3, 6, 9, 12, 15, 18\) give \(n^3 = 27, 216, 729, 1728, 3375, 5832\) — all less than 9999. That is 6 values.