Since a ball is drawn three times and the sum is $6$, the only ways this can happen are the permutations of $1, 2, 3$ and $2, 2, 2$. Therefore, with there being $3! + 1 = 7$ equally-likely possibilities in total, the probability that the ball numbered $2$ was drawn all three times is $\boxed{\textbf{(C)}\ \frac{1}{7}}$.
Three balls marked $1,2$ and $3$ are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is $6$, what is the probability that the ball numbered $2$ was drawn all three times? 
$\textbf{(A)} \ \frac{1}{27} \qquad  \textbf{(B)} \ \frac{1}{8} \qquad  \textbf{(C)} \ \frac{1}{7} \qquad  \textbf{(D)} \ \frac{1}{6} \qquad \textbf{(E)}\ \frac{1}{3}$