this is my attempt: Since $R$ is commutative, we let $I$ to be a prime ideal of $R$, the for $a,b\in R$,then the product $ab$ we must have that $a\in I$ or $b \in I$, by definition of a prime ideal. On the other hand,we have: $radI = \{r\in R | r^n\in I $ for some $n\in \mathbb{N}\}$. so the argument goes by induction on $(ab)^n$, for $n=1$ is just the definition of a prime ideal, so we assume that its true for $n$. now for $n+1$, we have the product: $(ab)^{n+1}= (ab)^n(ab)= (a^{n}a)(b^{n}b)$, but if $a\in I$, then $a^n\in I$, similarly if $b\in I$, then $b^n\in I$, therefore we thus have that $radI=I$. Now, i need to give an example that the converse is false, so if $a,b\in radI$, then $a^n\in I$ and $b^n\in I$, so when we consider the product $a^nb^n=(ab)^n$, we thus have that $a^n\in I$ and $b^n\in I$, therefore $I$ cant be a prime ideal.
My question is if this is the correct proof, especially in the converse.
If $R$ is a commutative ring and $P$ is a prime ideal of $R$, then $rad(P)=P$. To prove this fact, we first prove by induction over $n$ that if $x \in R$ and $n \in \mathbb N$, $n>0$ then $x^n \in P \Rightarrow x \in P$. If $n=1$, there is nothing to prove. Now suppose $n \ge 1$ and $x^n \in P \Rightarrow x \in P$ for every $x \in R$. If $x^{n+1}=xx^n \in P$ then either $x \in P$ or $x^n \in P$, but in the second case we still have $x \in P$.
Now let's prove the main assertion. If $x \in rad(P)$ then $x^n \in P$ for some $n \in \mathbb N$, $n>0$ by definition of radical. As we saw this imply $x \in P$, so we have $rad(P) \subset P$. The converse inclusion is obvious and holds for every ideal of $R$.
Conversely, if $R$ is a commutative ring and $I$ is an ideal of $R$ such that $rad(I)=I$, it is not necessary for $I$ to be prime. As a counterexample, consider $R= \mathbb Z$, and $I=(6)$. We have $rad(I)=I$ but $I$ is not prime.