Follow the hint. In the starting case, with probability $1/3$ it rains, the professor takes the umbrella, and with probability $2/3$, it does not rain when it is time for the professor to return home. So with probability $2/9$ the professor has not walked in the rain but the umbrella is at the office.
Similarly, with probability $1/9$, it has rained both to and from work and the umbrella has made a round trip.
With probability $2/9$, it did not rain on the way to work but it rained on the way back from work, meaning the professor got wet.
With probability $4/9$, it did not rain either way, and the professor is back home.
We can summarize this in a table for the round trip:
$$\begin{array}{ccccc}
\text{Umbrella} & \text{Rain} & \text{Got wet} & \text{Probability} \\
\hline
\text{Office} & \text{Yes, No} & \text{No} & 2/9 \\
\text{Home} & \text{Yes, Yes} & \text{No} & 1/9 \\
\text{Home} & \text{No, Yes} & \text{Yes} & 2/9 \\
\text{Home} & \text{No, No} & \text{No} & 4/9 \\
\end{array}$$
Therefore, with probability $5/9$, we have returned to the initial state (not wet, umbrella home), except a day has passed. Thus the expected number of additional days until getting wet is still $\mu$. With probability $2/9$, the professor got wet that day. With probability $2/9$, the professor has survived a day but now the umbrella is at the office. Since $v$ represents the expected number of days until getting wet when the professor is home but the umbrella is not, we summarize the expected number of days until getting wet is $$\mu = \frac{5}{9}(1 + \mu) + \frac{2}{9}(1) + \frac{2}{9}(1 + v).$$
Now for $v$, we suppose the professor begins the day at home but the umbrella is at the office. Then with probability $1/3$, the professor must walk in the rain to work. With probability $2/9$, the professor makes it to the office, and takes the umbrella home because it rains when it is time to leave. With probability $4/9$, it does not rain at all and the professor survives a day but returns to the state where the umbrella is not at home. So the expected number of days until getting wet in this case is...? I have not given the formula so that you have a chance to do the rest.
Ok, so it suddenly hit me that an umbrella, by itself, is not independent of the others.
Meaning, I cannot so simply say that the probability of a transition from home to office is $p$, because the probability of the transition is the probability that [it rains and this particular umbrella is chosen out of all the umbrellas that are in the same place]. This is the reason why my transition matrix is wrong, and making it correct would require too much work compared to the known solution (I would have to account for the case an umbrella is by itself, or with one other, etc.).
Best Answer
It would help a lot if you gave some indication as to how you got $0.19$ --- then maybe someone would see an arithmetical mistake, and we could all go home. Since you haven't done that, and since I'm not inclined to do the whole problem, I'll just indicate how one might get proceed.
$X_n$ goes up by one when it rains in the morning and not in the afternoon. The probability of this is $.08$ (unless $X_n=3$, in which case the probability is zero).
$X_n$ goes down by one when it doesn't rain in the morning but does rain in the afternoon. The probability of this is $.24$ (unless $X_n=0$, in which case the probability is zero).
This gives you all the transition probabilities; you can now figure out the eigenvector for the eigenvalue $1$ and thus the answer to the question.