Nice number

We know that if a number is of the form

$n = p_1^{a_1}\times p_2^{a_2} \times p_3^{a_3} .... \times p_n^{a_n}$ ... where $a_i$ s are primes,

then the number of it's factors is $(a_1+1)(a_2+1)(a_3+1)...(a_n+1)$ ... this is because in the factor, the power of prime $p_j$ could be anything from $0$ to $a_j$ implying $j+1$ different powers and talking about all primes together will give that number of factors.

Thus if you want a number to have 4 factors exactly, it HAS to be of the form $p^3$ where $p$ is prime, or of the form $p_1\times p_2$ where $p_1$ and $p_2$ are primes.

Here you have 5 choices of primes, and they are $2,3,5,7,11$ so you have to chose 2 out of them and have power 1. So there are $\dbinom{5}{2} =10$ numbers.

And other way, the cubes can only be $2^3$ and $5^3$ , so there will be 2 more numbers.

That accounts to give total $\boxed{12}$ numbers.