 Solution

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus