 Solution

If $x\in A$ is a unit in $B$, then $x^{-1}$ is integral over $A$:Multiplication by $x^{n-1}$ expresses $x^{-1}$ as an element of $A$, so $x$ is a unit in $A$.

Use the characterization $a\in J(A)$ exactly when $1-ra$ is a unit for every $r\in A$. If $a\in J(B)\cap A$, then $1-ra$ is a unit in $B$ and hence in $A$, proving $a\in J(A)$. Conversely, if $a\in J(A)$, every maximal ideal $\mathfrak{n}$ of $B$ contracts under the integral extension to a maximal ideal of $A$, which contains $a$. Thus every $\mathfrak{n}$ contains $a$, so $a\in J(B)$. Therefore