Hey,
For n^3 to be divisible by 24, n^3 should contain all the prime factors of 24. So now that we know that we need to prime factorize 24.
24 = 2^3 * 3
In any cube the number of prime factors are raised to a power of 3. For ex 8 = 2^3 , 64 = (2^2)^3
So we know that n should contain one 2 and a 3. Which means n has to be at least a 6.
However, we don’t know if n^3 has more than 3 2s. It may or it may not. For n to be 12 we need 6 2s in n^3. Hence we cannot say for sure.
The question is a “must be” question not a “could be question”. If it were a “could be” question we could have assumed n to be 12.
Hope this helps 