Let's break the problem down into two parts:

- If it takes $n$ rolls to get a $6$, how many times is $3$ likely to come up?
- How likely is it that it will take $n$ rolls to get a $6$?

For the first question, if a $6$ comes up for the first time on the $n$th roll, then one of the other $5$ numbers came up on each of the $n-1$ previous rolls. On each of those rolls the probability of getting a $3$ is $1/5$, so we would expect that $3$ will occur $\frac{n-1}{5}$ times.

For the second question, the probability that it takes $n$ rolls to get to the first occurrence of a $6$ is given by $(5/6)^{n-1} \cdot (1/6)$.

Now to answer the question: You can think of the expected number of $3$s as the sum of the expressions that answer the first question above, weighted by the probabilities that those situations occur. That is, we need to calculate

$$\frac{0}{5} \cdot (5/6)^0 \cdot (1/6) + \frac{1}{5} \cdot (5/6)^1 \cdot (1/6) + \frac{2}{5} \cdot (5/6)^2 \cdot (1/6) + \frac{3}{5} \cdot (5/6)^3 \cdot (1/6) +\dots $$

We can factor this a bit:

$$\frac{1}{30} \left( \frac{5}{6} + 2\left( \frac{5}{6} \right)^2 +3\left( \frac{5}{6} \right)^3 + 4\left( \frac{5}{6} \right)^4 + \cdots\right) $$

The sum in the parentheses is of the form $\sum_{n=0}^{\infty}na^n $, which converges to $\frac{a}{(1-a)^2}$; in this case we have $a=5/6$ so the sum in parentheses is $\frac{5/6}{(1/6)^2} = 30$, and therefore the whole expression is just $1$.

In fact the naive intuition (it should take about $6$ rolls to get to a $6$, and on each of the previous $5$ rolls there is a $1$ in $5$ chance of getting a $3$, so it should happen once) leads you to the correct answer, but I am not sure if that is just good fortune or if it reflects some deeper truth. I suspect the latter.