Let $X_i$ be a sequence of independent random variables with:
$$\Pr(X_i=3^i)=\frac{1}{2},\quad \Pr(X_i=-3^i)=\frac{1}{2}$$
Find $$R=\sup\{r\in \mathbb{R}:\Pr(|S_n|\geq r)=1\}$$ for $S_n=\sum_{i=1}^nX_i$. Find $\lim_n \frac{1}{n} R_n$.
We have $$\mathbb{E} X_i=\frac{1}{2}(3^i-3^i)=0,\quad \mathbb{V}(X_i)=\mathbb{E}X^2=3^{2i}\to \infty~(i\to\infty).$$
So the weak law of large numbers doesn't have to hold.
$$\Pr(|S_n|\geq r)=1 \iff \Pr(|S_n|<r)=0$$
Not sure where to go from here.
I presume you mean to write $$R_n = \sup \{ r \in \mathbb R : \Pr[|S_n| \ge r] = 1 \}.$$ Then for $n = 1$, $|S_1| = |X_1|$ and with probability $1$, this value is at least $3$, so $R_1 = 3$. So far, this is straightforward.
Now, for $n = 2$, what is the distribution of $|S_2|$? We observe the only permissible values are in $\{-12, -6, 6, 12\}$, and each occurs with equal probability, thus $|S_2| \ge 6$ with probability $1$ and $R_2 = 6$.
Let's try one more, $n = 3$: a similar computation shows $$S_3 \in \{\pm 15, \pm 21, \pm 33, \pm 39\},$$ and by now it should be clear that this is not even a question about probability; $R_3 = 15$ and we should be thinking more along the lines of discrete mathematics.
In the general case, $R_n = \min |S_n|$ is the smallest positive integer that can be represented by choosing coefficients $a_k \in \{\pm 1\}$ in the finite series $$\sum_{k=1}^n a_k 3^k.$$ We have the first few terms, but it doesn't seem clear how to get a closed form formula. Obviously, $3 \mid R_n$ for all $n$.
Without loss of generality, we can let $a_n = 1$. Then the first thing to test is if we choose $a_k = -1$ for all $k = 1, 2, \ldots, n-1$, will the resulting sum still be positive? If so, we have our solution, since it will not be possible to make a smaller sum. But note the geometric series $$\sum_{k=1}^{n-1} 3^k = \frac{3^n - 3}{2} < 3^n,$$ hence our minimum indeed occurs when $a_1 = \ldots = a_{n-1} = -1$, and $a_n = 1$, and this minimum value is simply $$R_n = 3^n - \frac{3^n - 3}{2} = \frac{3^n + 3}{2}.$$ All that remains is to compute $$\lim_{n \to \infty }\frac{3^n + 3}{2n}.$$