It has got to be one of these 4 numbers, right?

The product of the positive divisors of positive integers $N$ is given by $\begin{aligned} &N^{\frac{d(N)}{2}} = 6\times 16\times 36\times 96 \\& N^{\frac{d(N)}{2}} =(3\times 2).(2^4).(3^2\times 2^2).(2^5\times3 ) \\& N^{\frac{d(N)}{2}} = (2^3\times 3)^{\frac{8}{2}}\implies N=\boxed{24}\end{aligned}$ Where $d(N)$ is the number of positive divisors of $N$ and it's true that $24$ has $8$ positive divisors ie $d(24) =8$ so the answer is $24$ .

Since only powers of two and three appear, $N$ is 3-smooth (otherwise it's other prime factors would show up in the product). So $N = 2^p 3^q$ . Then, the product of the prime factors of N is $2^a3^b$ , where $(a,b)$ is the sum of all ordered pairs $(x,y)$ where $(0 \leq x \leq p)$ and $(0 \leq y \leq q)$ . This turns out to be: $((q+1)\Delta(p),(p+1)\Delta(q))$ , where $\Delta(x)$ is the sum of all positive integers not greater than x. Equating this to $(12, 4)$ and trying out the possible factorizations yields only $(a,b) = ( 2(1+2+3) , 4(1))$ , so $p = 3$ and $q = 1$ . Then, $N = 2^33^1 = 24$ .