Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$
This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.
Theorem: $\;\mathbb Z_{\large mn}\;$ is cyclic and $$\mathbb Z_{\large mn} \cong \mathbb Z_{\large m} \times \mathbb Z_{\large n}$$
if and only if $\;\;\gcd(m, n) = 1.$
This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$
It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.
Good-to-know Corollary/Generalization:
The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and
$$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.
The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are
$$\prod_i P(e_i)$$
non-isomorphic Abelian groups of order $n$. You missed $2+2$ among the ways to partition $4$, corresponding to $\mathbb Z/(2^2\mathbb Z) \times \mathbb Z/(2^2\mathbb Z) = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$.
You can use this formula to answer the original question. The first few partition numbers are:
P(1) = 1 1
P(2) = 2 2, 1+1
P(3) = 3 3, 2+1, 1+1+1
P(4) = 5 4, 3+1, 2+2, 2+1+1, 1+1+1+1
P(5) = 7 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1
and they obviously increase from here. Now, if there are $4$ Abelian groups of order $n$, then we must have $e_1=e_2=2$, and we choose as bases the two smallest primes. Thus the number we are looking for is $2^23^2=36$.
Best Answer
You can use the fundamental theorem of finitely generated abelian groups, which states that each of your groups can be expressed uniquely (up to isomorphism) as the product cycles of order the power of primes and compare factors and the prime powers of the cycles of any two such decompositions to determine whether they are isomorphic, up to the order of their factors.
For example, let's take $\mathbb Z_8 \times \mathbb Z_{10} \times \mathbb Z_{24}$, and decomposing this gives us:
$$\mathbb Z_8 \times \color{blue}{\bf \mathbb Z_{10}} \times \color{red}{\bf \mathbb Z_{24}} \quad \cong \quad \mathbb Z_8 \times \color{blue}{\bf (\mathbb Z_{2} \times \mathbb Z_5)} \times \color{red}{\bf (\mathbb Z_{3}\times \mathbb Z_{8})} \cong {\bf \mathbb Z_{2} \times \mathbb Z_{3}\times \mathbb Z_5 \times \mathbb Z_{8}^2}\tag{1}$$
Now compare this to the decomposition you get for the expression you are to compare with $(1)$: $$\Bbb Z_4 \times \Bbb Z_{12} \times \Bbb Z_{40} \quad \cong \quad \mathbb Z_4 \times (\mathbb Z_3 \times \mathbb Z_4)\times (\mathbb Z_8\times \mathbb Z_5) \cong {\bf \mathbb Z_3 \times \mathbb Z_4^2 \times \mathbb Z_5 \times \mathbb Z_8}\tag{2}$$
Note that there the cyclic factors of $(1), (2)$ are not equivalent, up to the order of their appearance, hence not isomorphic.