= Solution
Use the following <Jacobson radical> facts for a unital algebra: a <nilpotent ideal> is contained in the <Jacobson radical>; under a surjective algebra homomorphism the image of the <Jacobson radical> is contained in the radical of the quotient; and a finite product of fields has zero <Jacobson radical>. For completeness, if $I^N=0$ and $x\in I$, then $ax\in I$ for every $a\in A$, and $1-ax$ is invertible with inverse $\sum_{j=0}^{N-1}(ax)^j$. The usual unit criterion for the <Jacobson radical> therefore gives $I\subseteq J(A)$. The quotient assumption gives $J(A)\subseteq I$. Hence
$$
\boxed{I=J(A).}
$$
This is the <nilpotent ideal with semisimple quotient radical criterion>.
For the finite <quiver> under consideration, let $R$ be the <arrow ideal of a path algebra>, spanned by paths of positive length. If $Q$ has $r$ vertices and no oriented cycle, a path cannot repeat a vertex, so $R^r=0$. Meanwhile $kQ/R\cong\prod_{i\in Q_0}k$, with the constant paths giving the coordinate idempotents. The criterion just proved yields \b[$J(kQ)=R$].
The condition on cycles is necessary. A <quiver with one loop> has <path algebra> $k[t]$, whose arrow ideal is $(t)$. But \b[$J(k[t])=0\ne(t)$]: the maximal ideals $(t-a)$, $a\in k$, have intersection zero, since the algebraically closed field $k$ is infinite and a nonzero polynomial has only finitely many roots. Thus the arrow ideal need not be the <Jacobson radical> when oriented cycles are present.
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