Expected number of steps between states in a Markov Chain
The distribution for the number of time steps to move between marked states in a discrete time Markov chain is the discrete phase-type distribution. You made a mistake in reorganising the row and column vectors and your transient matrix should be $$\mathbf{Q}= \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{2}{3} & 0 & \frac{1}{3}\\ \frac{2}{3} & 0 & 0 \end{bmatrix}$$ which you can then continue to find $$\mathbf M = (\mathbf I - \mathbf Q)^{-1} = \begin{bmatrix} 27 & 9 & 3\\ 24 & 9 & 3 \\ 18 & 6 & 3\end{bmatrix}$$ and summing the first row gives you (as you require) 39.
Generally when I have seen these distributions written the 'exit' state is put last, which would mean your matrix with elements in the order $\{0,1,2,3\}$ was already in canonical form and you needed to extract the top left portion as your transient matrix,
$$\mathbf P = \left( \begin{array}{ccc|c} \frac{2}{3} & \frac{1}{3} & 0 & 0\\ \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ \frac{2}{3} & 0 & 0 & \frac{1}{3}\\ \hline\\ 0 & 0 & 0 & 1 \end{array} \right) $$ For further moments you might be interested in this paper:
- Dayar, Tuǧrul. "On moments of discrete phase-type distributions." Formal Techniques for Computer Systems and Business Processes. Springer Berlin Heidelberg, 2005. 51-63. doi:10.1007/11549970_5.
Usually, my approach to this kind of question is to solve a very simple recurrence.
Just look to $ E_{0} N(3)$ but conditioned in each of the two possible first steps to get (I will write $N$ instead of $N(3)$
$$E_0 N = \frac{2}{3}E_0N + \frac{1}{3}E_1N +1$$
The term +1 shows up because once you had walked one step you have to add this step to the account. (if you want a rigorous proof of this equation, use the simple property of Markov)
If you repeat this process to $E_1N$ and $E_2N$ and replace what you find in the first equation you will find that $E_0N = 39$ :(
Now I'm curious about the results... I made the calculations twice, but it is possible that I forgot something.