An even integer $2n$ is prime in the $\Bbb{E}$-zone if and only if $2n \equiv 2 \pmod{4}$, which occurs if and only if $n$ is odd. So, factorization into primes amounts to writing
$$
\begin{align}
N &= 2n_1 \cdot 2n_2 \cdots 2n_k \\
&= 2^k \cdot n_1 \cdot n_2 \cdots n_k,
\end{align}
$$
where $n_1, \ldots, n_k$ are odd integers. Since we're looking for small numbers with many prime factorizations, and since the product of odd numbers is odd, we may as well assume that $k = 2$. (You seem to have done this implicitly by writing prime factorizations as pairs.)
If we want $4$ prime factorizations, then we're looking for $4$ pairs of prime factors, or an odd number with $8$ factors. (The factors are automatically odd, too.)
In the integers, the number of factors of the number $p_1^{e_1} \cdots p_r^{e_r}$, where $p_1, \ldots, p_r$ are usual primes, is
$$
(1 + e_1) \cdots (1 + e_r).
$$
So, we need to write $8$ (the number of factors we desire) as $2 \cdot 2 \cdot 2$ or as $4 \cdot 2$. We will consider each possibility in turn.
In the first case, we must consider a product of distinct (odd!) primes $p_1 \cdot p_2 \cdot p_3$. The smallest example of this is $3 \cdot 5 \cdot 7 = 105$, whose factors can be paired off as:
$$
\begin{array}{ccc}
1 && 3 \cdot 5 \cdot 7 \\
3 && 5 \cdot 7 \\
5 && 3 \cdot 7 \\
7 && 3 \cdot 5
\end{array}
$$
Back in the $\Bbb{E}$-zone, these are prime factorizations of
$$
2^2 \cdot 3 \cdot 5 \cdot 7 = 420
$$
(by putting a $2$ on each factor of the pair):
$$
\begin{array}{ccc}
2 && 210 \\
6 && 70 \\
10 && 42 \\
14 && 30
\end{array}
$$
In this case, consider a number of the form $p_1^3 \cdot p_2$. The smallest example of this is $3^3 \cdot 5 = 135$, which is larger than $105$, but I'll follow through the analysis anyway. The factors in pairs are:
$$
\begin{array}{ccc}
1 && 3^3 \cdot 5\\
3 && 3^2 \cdot 5 \\
3^2 && 3 \cdot 5 \\
3^3 && 5
\end{array}
$$
Back in the $\Bbb{E}$-zone, these are prime factorizations of
$$
2^2 \cdot 3^3 \cdot 5= 540
$$
(by putting a $2$ on each factor of the pair):
$$
\begin{array}{ccc}
2 && 270 \\
6 && 90 \\
18 && 30 \\
54 && 10
\end{array}
$$
So, $420$ is the smallest positive integer with $4$ prime factorizations in the $\Bbb{E}$-zone.
This follows from properties of the divisor function. Say we have some number $n$ and we want to calculate how many divisors it has. This is the same as calculating $\sigma_0(n)$, where
$$\sigma_0(n)=\sum_{d|n}1.$$
If we write $n=p_1^{a_1}\cdots p_k^{a_k}$ as a product of primes, then one may show that we have
$$\sigma_0(n)=\prod_{i=1}^k(a_i+1).$$
You want to find the least $n$ such that $\sigma_0(n)=60$. So let's factorize 60 into primes. As you noted, this is $2\times 2\times 3\times 5$. Now it just remains for you to argue that we must have $k=4$, and $\{a_1+1,\dots,a_4+1\}=\{2,2,3,5\}$ under some ordering, then say why the choice you give in the question is minimal.
Best Answer
Every divisor of an odd number is odd, su just skip the prime factor $2$. That is, if you want to find a number with $N$ odd divisors, where $$N=p_1\cdot\ldots\cdot p_s$$ and $$p_1\ge\ldots\ge p_s$$ are prime numbers, not necessarily different, take $$3^{p_1-1}\cdot 5^{p_2-1}\cdot\ldots\cdot q_s^{p_s-1}$$ where $q_s$ is the $s$-th odd prime number.
For your example ($N=12$). Since $N=3\cdot2\cdot 2$, the number you re looking for is $3^2\cdot 5\cdot 7=315$.