For any
$$n\in\Bbb N\;,\;\;\text{ord}\,\left(\frac1n\right)_{\Bbb Q/\Bbb Z}=n$$
So we already know there's a cyclic subgroup of order $\,n\,$ in $\,\Bbb Q/\,\Bbb Z\,$ . Now, if
$$\left(\frac ab+\Bbb Z\in\Bbb Q/\Bbb Z\;\;\;\text{and}\;\;\;\text{ord}\,\left(\frac ab\right)_{\Bbb Q/\Bbb Z}=n\right)\implies \left(n\frac ab\in\Bbb Z\right)\iff \left(n=bk\;,\;k\in\Bbb Z\right)$$
and thus in fact we have that
$$\frac ab=\frac{ak}n\in\left\langle\;\frac1n+\Bbb Z\;\right\rangle\le\Bbb Q/\Bbb Z$$
and this gives us uniqueness
The correct condition is that $m$ and $n$ are coprime, that is, have no common factors. In particular, this means that $C_2 \times C_3 \cong C_6$. (Sometimes it is hard to compare two multiplication tables by inspection, though it can help identifying two isomorphic groups from their tables by reordering the elements.)
Hint Given two groups $G, H$ and elements $a \in G, b \in H$, the order of element $(a, b)$ is the smallest number, $\text{lcm}(k, l)$ divisible by both the order of $l$ of $a$ and the order $k$ of $b$.
In particular, suppose $G$ and $H$ are cyclic, say $G = C_n$, $H = C_m$. Then, if $a$ and $b$ are generators of $C_n$, $C_m$, respectively, by definition they have respective orders $n, m$ and hence $(a, b) \in C_n \times C_m$ has order $\text{lcm}(m, n)$.
So, if $m, n$ are coprime, this order is $\text{lcm}(m, n) = mn$.
Conversely, if $m, n$ are not coprime, then since any $a \in C_n$ has order $l$ dividing $n$ and any $b \in C_m$ has order $k$ dividing $m$, we have that the generic element $(a, b) \in C_n \times C_m$ has order $\text{lcm}(k, l) \leq \text{lcm}(m, n) < mn$. In particular, no element has order $mn$, so $C_m \times C_n$ is not cyclic.
For example, the elements $1 \in C_2$ and $1 \in C_3$ generate their respective groups, and the powers of $(1, 1) \in C_2 \times C_3$ are
$$(1, 1), (0, 2), (1, 0), (0, 1), (1, 2), (0, 0),$$
so $(1, 1)$ generates all of $C_2 \times C_3$, which is hence isomorphic to $C_6$.
Best Answer
You can compute the center and you will get that it is a group of order $2$ such that the quotient has order $4$ as you mentioned. Therefore C) or D) could be correct by now. If you are considering the elements in the quotient you will see that every non-trivial element has order $2$ and therefore you will get the Klein-$4$-group. So yes, check whether the quotient is cyclic by computing the orders.
Let me give an example. The order of $i \cdot \lbrace 1,-1 \rbrace \in Q_8/Z(Q_8)$ is $2$, as $i \not\in \lbrace 1,-1 \rbrace$ and
$(i \cdot \lbrace 1,-1 \rbrace)^2 = i^2 \cdot \lbrace 1,-1 \rbrace = -1 \cdot\lbrace 1,-1 \rbrace = 1 \cdot \lbrace 1,-1 \rbrace$.
Now do the computation for the other $2$ non-trivial elements of the quotient.