The answer is $\boxed{\text{D}}$.
The given series is $3, 9, 15, \dots$. This is an arithmetic series with a common difference of $6$. The general formula for an arithmetic series is $a_n = a_1 + d(n-1)$, where $a_n$ is the $n$th term, $a_1$ is the first term, $d$ is the common difference, and $n$ is the number of terms.
In this case, we know that $a_1 = 3$ and $d = 6$. We want to find $a_{21}$. So we have:
$$a_{21} = a_1 + d(21-1) = 3 + 6(20) = 129$$
Therefore, the 21st term of the series is $\boxed{129}$.
Option A is incorrect because $117$ is not a term in the series. Option B is incorrect because $121$ is not a term in the series. Option C is incorrect because $123$ is not a term in the series.