This is called "first-transition analysis." Let $h\left(i\right)$ be the mean time to reach state $1$ given that you start in state $i$ (you can then repeat this same procedure for the other two target states). Then,
h\left(1\right) &=& 0,\\
h\left(2\right) &=& 1 + 0.5h\left(1\right) + 0.5h\left(3\right),\\
h\left(3\right) &=& 1 + h\left(2\right).
If you are already in state $1$, you have already reached it, so the mean time is zero. If you are in state $2$, you first spend one time unit to transition out of state $2$ (that is the significance of adding $1$ in front). After that you will be in either state $1$ or state $3$, and once you are in either of those, the problem "restarts" with that as your starting state -- since this is a Markov chain, the distribution of the future trajectory depends only on the most recent known history. If the problem restarts in state $3$, the expected time to reach $1$ from that point on will be $h\left(3\right)$ by definition, so you just need to take a weighted average of $h\left(1\right)$ and $h\left(3\right)$ because your next state (the one used to restart the problem) will actually be randomly determined.
The above is a system of linear equations, so you can easily solve it. Now, what you actually want to compute is the mean return time to state $1$ given that you started there. From the transition matrix, this is equal to $1 + h\left(2\right)$, since you spend one time unit to leave state $1$, and after that you will automatically "restart" in state $2$.