Let $G$ be a group and $|Z(G)| = 1$. By $N(G)$ we denote the set of orders of conjugacy classes of the group $G$.
If $|G| = 60$ and $N(G) = \lbrace 1, 12, 15, 20 \rbrace$ then the normalizer of the Sylow $2$-subgroup has order $12$.
I tried to prove it myself, but I'm afraid there may be mistakes in my proof.
$\textbf{Proof:}$
Denote the normalizer of the Sylow $2$-subgroup by $N$.
The number of Sylow $2$-subgroups is equal to $|G:N|$. On the other hand, we understand that $N$ is a subgroup of $G$, and $S\in Syl_2(G)$ is a subgroup of $N$.
Given these facts, we get that $|N|$ is divisible by $4$ and divides $60$. So either $|N| = 4$, or $|N| = 12$, or $|N| = 20$, or $|N| = 60$.
Let's consider the case when $|N| = 20$. But then it's not difficult to understand that we will get a contradiction with the $\textbf{Index theorem}$ (A group $G$ with a subgroup $H$ of index $n$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$). It's enough just to sort out the possible divisors of $20$ and apply the specified theorem.
If $|N| = 60$ then the number of Sylow $2$-subgroups is equal to $1$. We know that $\exists x\in G\;|\;|C_G(x)| = 4$ (since one of the conjugacy classes has an order equal to $12$).
$C_G(x)$ is then the only Sylow $2$-subgroup, which means it is characteristic (in particular, it is normal) in $G$.
Let $g\in C_G(x)$. By definition of a normal group, we have that $g^h\in C_G(x)$ for any $h\in G$. Hence, $C_G(x)$ consists of several conjugacy classes. But due to the existing $N(G)$, we get that $C_G(x)$ can consist of only $4$ different elements of $y$, such that $y^{h} = y\;\forall h\in G$. But then $y\in Z(G) \Rightarrow |Z(G)| > 1.$ We get a contradiction because by the condition $|Z(G)| = 1$.
By similar reasoning, we can understand that $n_3(G) >1$ and $n_5(G)>1$.
Consider the case when $|N| = 4$.
Let's try to understand what the minimum number of Sylow $3$ and $5$-subgroups can be.
Based on Sylow's theorems, we can immediately say that $n_5(G) = 6$ and $n_3(G) = 10$ or $n_3(G) = 4$. However, $n_3(G) \neq 4$, because using $\textbf{Index theorem}$ , we would get that $|G/K| \; | \; 4! = 24$ (where $K$ is a normal subgroup of $G$, satisfying the condition of the theorem). It can be only if $|K| = 5$, but this is not the case due to the fact that the Sylow $5$-subgroup of $G$ is not normal.
Consider the Sylow $5$-subgroups of $G$. It is clear that they all intersect trivially $\Rightarrow$ the union of various nonunit elements of the Sylow $5$-subgroups is equal to $6\cdot 5 - 6 = 24$.
Similarly, we obtain that the union of various nonunit elements of the Sylow $3$-subgroups is equal to $20$.
In total, taking into account the unit, we get that the various elements from the Sylow $3$ and $5$-subgroups in the group $G$ are $45$. Then, given that $n_2(G) = 15$, we come to the fact that $\forall\;S, T\in Syl_2(G) \Rightarrow |S\cap T| = 3.$ But this cannot be, since $S\cap T\leq S$ and $3\nmid 4$ (contradiction with Lagrange's theorem).
So, we get that $|N| = 12$.
