Necessary and sufficient condition for $r(\mathfrak a)$ to be prime As we know, $$\mathfrak a~\text{is a primary ideal}\Rightarrow r(\mathfrak a)~\text{is a prime ideal}. $$ 
But $r(\mathfrak a)$ may not be a prime ideal if $\mathfrak a$ isn't a primary ideal.

What's the necessity and sufficiency condition of $r(\mathfrak a)$ is prime? primary?

 A: An ideal is prime iff it is primary and radical (aka semiprime.)
If we only talk about radical ideals, then such an ideal is prime iff it is primary. Remember that radical ideals have the property that $x^n\in I$ implies $x\in I$.
Of course, if $r(\mathfrak{a})$ is prime, then it's primary. 
In the other direction, suppose $r(\mathfrak{a})$ is primary. If $xy\in r(\mathfrak{a})$, then either $x\in r(\mathfrak{a})$ or else $y^n\in r(\mathfrak{a})$ for some $n$. We must show if $x\notin r(\mathfrak{a})$, then $y\in r(\mathfrak{a})$. Now, $y^n\in r(\mathfrak{a})$ implies $y\in r(\mathfrak{a})$. This shows that at least one of $x,y$ is in $r(\mathfrak{a})$ if $xy$ is in $r(\mathfrak{a})$, verifying that $r(\mathfrak{a})$ is prime.

(Added to answer more of your comment below.)
If $\mathfrak{a}$ is primary, then $r(\mathfrak{a})$ is prime. This is easy to prove from definitions: just start with $xy\in r(\mathfrak{a})$ and follow your nose through the definition of "primary." Of course, don't forget the characterization of $r(\mathfrak{a})$ given above.
The converse is not true. If you take a field $F$ and look at the ideal $I=(xy,y^2)$, you can easily check that any prime ideal containing $I$ contains the prime ideal $(y)$, so $r(I)=(y)$. And yet $I$ is not primary since $y\notin I$ and no power of $x$ is in $I$.
