Every finite ring $R$ has a nilpotent ideal $I$ such that the only nilpotent of $R/I$ is the zero ideal Just like the title.
How to show that every finite ring $R$ has a nilpotent ideal $I$ such that the only nilpotent ideal of $R/I$ is the zero ideal?
I just know that every finite ring has a nilpotent ideal, but how to construct a ideal $I$ satisfiying the additional condition?
I guess that the $I$ need to be large enough, but what's more?
 A: Consider a nilpotent ideal $I$ whose cardinal is maximal. If $R/I$ has a non trivial nilpotent ideal $J$, $p^{-1}(J)$ is nilpotent, where $p:R\rightarrow R/I$ is the quotient map.
A: 
I just know that every finite ring has a nilpotent ideal,

Well... every ring has the trivial ideal as a nilpotent ideal. You could not possibly have meant "nonzero nilpotent ideal" because obviously the finite fields don't have such an ideal, and are quite finite.

but how to construct a ideal  satisfiying the additional condition?

Finiteness plays a very ham-handed role, one which could easily be replaced by a weaker condition.
Consider for a moment any two ideals $I\subseteq J$ of a ring $R$. If the ideal $J/I$ is nilpotent in $R/I$, it means $J^k\subseteq I$ for some $k$.  If $I$ is also a nilpotent ideal of $R$, then so is $J^k$ and also $J$.
What this says is that in order for $R/I$ to lack nonzero nilpotent ideals, you want to find an $I$ that is maximal among nilpotent ideals of $R$.. That is how you will construct/find your answer.
Now... how you get that maximal nilpotent ideal is easily accomplished for a finite ring: there are only finitely many ideals so you just write them all down and pick a maximal nilpotent one.
But really all you needed is the existence of a maximal member of the poset of nilpotent ideals, and that would be given to you merely by the ring being right or left Noetherian.
A final fact worth mentioning is that for a right or left Artinian ring, there is a unique maximal nilpotent ideal: it is the Jacobson radical. The Jacobson radical always contains nilpotent ideals, but for Artinian rings in particular you know that the radical itself is nilpotent, and that makes it the biggest.
All of that last paragraph holds in particular for finite rings, so now you know exactly what ideal to look for.
