Formula for number of conjugacy classes in terms of characters [duplicate]

You have the correct interpretation of the question: the $K_i$ is a basis of the centre of the group algebra, and the $a_{ijv}$ would be called the "structure constants" for multiplication, with respect to this basis.

We need to show that the $a_{ijv}$ may be found with the following formula: $$ |G| a_{ijv} = |C_i| |C_j| \sum_V \frac{1}{\dim V} \chi_V (g_i) \chi_V (g_j) \overline{\chi_V (g_v)}$$ where the sum is over all irreducible representations $V$ of $G$, and $\chi_V$ denotes the (irreducible) character associated to the irreducible representation $V$.

Firstly, by column-orthogonality of characters, we have that $$\sum_V \chi_V(g_i) \overline{\chi_V(g_j)} = \delta_{ij} \frac{|G|}{|C_j|}.$$ So, fix a conjugacy class $C_v$ and apply $f_v(x) = \sum_V \chi_V(x) \overline{\chi_V(g_v)}$ to both sides of $K_i K_j = \sum_w a_{ijw} K_w$ to get $$ \sum_V \chi_V(K_i K_j) \overline{\chi_V(g_v)} = a_{ijv} |G| $$

From here we just have to show that the left hand side is what we want, which we can do by considering each term of the sum individually. Fix an irreducible representation $V$. Since $K_i$ and $K_j$ are elements of the centre of the group algebra, they will each act on $V$ by a scalar, so we may write $\chi_V(K_i) = \lambda_i (\dim V)$, $\chi_V(K_j) = \lambda_j (\dim V)$ for some $\lambda_i, \lambda_j \in \mathbb{C}$. Crucially, we also have that $$\chi_V(K_i K_j) = \lambda_i \lambda_j (\dim V) = \frac{1}{\dim V} \chi_V(K_i) \chi_V(K_j).$$

So now we have $$\sum_V \frac{1}{\dim V}\chi_V(K_i) \chi_V(K_j) \overline{\chi_V(g_v)} = a_{ijv} |G|$$

Finally, since characters are constant on conjugacy classes, we have $\chi_V(K_i) = |C_i| \chi_V(g_i)$ and similarly for $\chi_V(K_j)$, and so we are done.