 Solution

Put $K={\bigcap }_{L}L$, where $L$ ranges over all maximal left ideals. For each such $L$, the action on the simple module $A/L$ has a primitive ideal $P_L$ as its kernel. An element of that kernel annihilates $1+L$, so $P_L\subseteq L$. The intersection $J$ of all primitive ideals is consequently contained in every $L$, hence $J\subseteq K$.

Conversely, take $x\in K$ and any nonzero simple module $E$. For every nonzero $v\in E$, simplicity makes the evaluation map $A\to E$, $a\mapsto av$, surjective. Its kernel is a maximal left ideal, so it contains $x$. Thus $xv=0$ for every nonzero $v$, and also for zero. The element $x$ therefore belongs to the annihilator of every simple module, hence to every primitive ideal. This provesSince the primitive ideals are two-sided ideals, this argument also makes $J$ a two-sided ideal, despite the one-sided ideals in this alternative description.