As an example of a commutative algebra object which can also be turned into a Frobenius algebra, consider the algebra of polynomials $\mathbb{C}[X]$ in one indeterminate $X$ over the field $\mathbb{C}$, divided by the ideal $\langle X^d \rangle$, i.e. $A=\mathbb{C}[X]/\langle X^d \rangle$. This is a commutative algebra object in the category of finite dimensional complex vector spaces $\mathbf{Vect}_{\mathbb{C}}$. The reason one divides by the ideal $\langle X^d \rangle$ is to make the algebra object (viewed as a vector space over $\mathbb{C}$) finite dimensional, in order to be able to turn it into a Frobenius algebra. Thus in the case at hand $\dim A=d$. The tensor product bifunctor $\otimes_{\mathbb{C}} \colon \mathbf{Vect}_{\mathbb{C}} \times \mathbf{Vect}_{\mathbb{C}} \to \mathbf{Vect}_{\mathbb{C}}$ is given by $\cdot \colon A \times A \to A$. More explicitly the multiplication is
$$
\left( \sum_{i=0}^{d-1} a_i X^i \right)\cdot \left( \sum_{j=0}^{d-1} b_j X^j \right) = \sum_{k=0}^{d-1}\left( \sum_{i+j=k} a_ib_j \right)X^k ~,
$$
where $a_i,b_j\in \mathbb{C}$ and $X^i\in A$. The monoidal unit is given by the underlying field $\mathbb{C}$.