Assume $A$ is a connected (not necessarily commutative) non-semisimple Frobenius algebra that is finite dimensional over a field of characteristic 0. (for the commutative case being connected means it is local).
We should have $c(1)= \sum_{i} y_i \otimes x_i$, where $x_i, y_i$ for $i=1,...,dim A$ are defined by the condition $b (x_i y_j )= \delta_{i,j}$ (the existence of such $x_i, y_i$ is equivaleng to $A$ being a Frobenius algebra.)
Now $\mu c(1)= \sum y_i x_i$ and $b (\mu c(1))= \sum b(y_i x_i)=dim A$. This implies that $\mu c(1) /dim A$ is in the socle and thus squares to zero.
Thus the statement is true in the case of commutative local Frobenius algebras when the algebra is not a division ring.
In the non-local case it should be wrong by taking for example $K \times K[x]/(x^2)$ as suggested in the comments by Will Sawin.