If $I=\langle 12 \rangle$, then $\text{Rad}(I)=\langle 6\rangle$ To show that if $I=\langle 12 \rangle$, then $\text{Rad}(I)=\langle 6\rangle$, I did the following: 
$$36=3 \cdot 12 \\ 6^2=36 \in I \Rightarrow 6 \in \text{Rad}(I) \Rightarrow \langle 6 \rangle \subseteq \text{Rad}(I)$$ 
Let $a \in \text{Rad}(I) \Rightarrow \exists n \in \mathbb{N}$ such that $a^n \in \langle 12 \rangle$ that means that $a^n=12 l, l \in \mathbb{Z} \Rightarrow 12 \mid a^n \Rightarrow 6 \mid a^n$ 
But how can we conclude from here that $a \in \langle 6 \rangle$ ?
 A: For any commutative ring $R$ and an ideal $\mathfrak a$, the radical of $\mathfrak a$ is the intersection of all prime ideals lying over $\mathfrak a$. In your case, the primes over $(12)$ are $(3)$ and $(2)$ (why?), and so $\mathfrak r(12)=(3)\cap (2)=([2,3])=(6)$. 
A: Lemma: Let $R$ be an integral domain, if $p, q$ are two primes in $R$ that are not associates (that is, $p \nmid q$ and $q \nmid p$) such that $p \mid x$ and $q \mid x$, then $pq \mid x$.
Proof: Since $p \mid x$, then $x = ps$ for some $s \in R$. Since $q \mid x = ps$ it follows that $q \mid s$ since $q$ is prime and $q \nmid p$ by hypothesis. Hence $s = qt$ for some $t \in R$. Thus $x = ps = p(qt) = (pq)t$ and thus $pq \mid x$ as desired. $\square$.
Claim: If $I = \langle 12 \rangle$, then $\text{rad}(I) = \langle 6 \rangle$.
Proof: Let $x \in \langle 6 \rangle$, then $x = 6 \cdot s$ for some $s \in \mathbb Z$. Observe that $$x^2 = (6s)^2 = 36s^2 = 12 \cdot (3s^2) \in \langle 12 \rangle.$$ Thus by definition $x \in \text{rad}(I)$.
Let $x \in \text{rad}(I)$, then there exists $n > 0$ such that $x^n \in I$, that is, $x^n = 12 \cdot m = 2^2 \cdot 3 \cdot m$ for some $m \in \mathbb Z$. In particular we have $2 \mid x^n$ and $3 \mid x^n$. Since both are prime we have $2 \mid x$ and $3 \mid x$. Since they are not associate, we can conclude by the lemma that $2 \cdot 3 = 6 \mid x$ and hence $x \in \langle 6 \rangle$.$\square$.
We can prove a slight generalization:
Claim: Let $R$ be a UFD, $r \in R$ with prime factorization $r = p_1^{k_1} \cdots p_n^{k_n}$, then $\text{rad}\langle r \rangle = \langle p_1 \cdots p_n \rangle$.
Proof: Let $x \in \langle p_1 \cdots p_n \rangle$, then $x = (p_1 \cdots p_n) \cdot s$ for some $s \in R$. Choose $k = \max\{k_1, \ldots, k_n\}$ and observe that $$x^k = (p_1 \cdots p_n)^k \cdot s^k = (p_1^k \cdots p_n^k) \cdot s^k$$ which means $g \mid x^k \Leftrightarrow x^k \in \langle g \rangle$ and hence $x \in \text{rad}\langle g \rangle$.
Let $x \in \text{rad}(g)$, then $x^k \in \langle g \rangle$ for some $k > 0$. Observe that $x^k = (p_1^{k_1} \cdots p_n^{k_n}) \cdot s$ for some $s \in R$. Notice that $p_1 \mid x^k, \ldots, p_n \mid x^k$ and hence we get $p_1 \mid x, \ldots, p_n \mid x$ by the primality of the $p_i$'s. Since $p_i \nsim p_j$ for $i \neq j$ (that is, they are not associates of each other), we can use a generalized version of the lemma above (easily shown using induction) and conclude that $p_1 \cdots p_n \mid x$. Conclude that $x \in \langle p_1 \cdots p_n \rangle$.
Note: $R$ is restricted to a UFD so that $r$ can be written as $r = p_1^{k_1} \cdots p_n^{k_n}$ uniquely up to units.
