= Solution
This is the <Artin--Tate lemma>. Choose $A$-algebra generators $x_1,\ldots,x_r$ of $C$ and $B$-module generators $e_1,\ldots,e_s$ of $C$. Write
$$
x_i=\sum_j a_{ij}e_j,\qquad
e_ie_j=\sum_\ell b_{ij\ell}e_\ell
$$
with coefficients $a_{ij},b_{ij\ell}\in B$, and let $B_0$ be the $A$-subalgebra of $B$ generated by these finitely many coefficients.
The module $\sum_jB_0e_j$ contains the $x_i$, is closed under multiplication, and contains $A$ after including an expression for $1$ among the chosen coefficients. It therefore equals $C$. Thus $C$ is a finite $B_0$-module. The ring $B_0$ is Noetherian by the <Hilbert basis theorem>, and $B\subseteq C$ is a $B_0$-submodule, so $B$ is a finite $B_0$-module. It follows that $B$ is a finitely generated $A$-algebra.
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