Notice, all days can be classified into one of two disjoint options:
Rainy, in which case we can deduce from the given info that P(Rain) = 5/365
Not Rainy, and since this is the complement of above, P(Not Rain) = 360/365
\begin{equation*}
P(\text{Rain}) = P(S_1) = \frac{5}{365}
\end{equation*}
and
\begin{equation*}
P(\text{Not Rain}) = P(S_2) = \frac{360}{365}.
\end{equation*}
Further, you are given the conditional probabilities
\begin{equation*}
P(\text{ Forecast Rain | Rain}) = P( A | S_1) = 0.9
\end{equation*}
\begin{equation*}
P(\text{ Forecast Rain | Not Rain}) = P( A | S_2) = 0.1
\end{equation*}
Notice that the question provided requests that you find the probability of Rain given that the weatherman has forecasted rain. What is given on the other hand is the reverse of that conditional probability. Using Bayes’ Theorem allows you to turn this around...
\begin{align*}
P(\text{Rain}) & = P(S_1) P( A | S_1) + P(S_2) P(A | S_2)\\
& = \frac{5}{365} \cdot 0.9 + \frac{360}{365} \cdot 0.1
\end{align*}
Hence, putting these together gives
\begin{align*}
P(\text{Rain | Forecast Rain}) & = \frac{\frac{5}{365} \cdot 0.9}{\frac{5}{365} \cdot 0.9 + \frac{360}{365} \cdot 0.1}\\
& = \frac{5 \cdot 0.9}{5 \cdot 0.9 + 360 \cdot 0.1}\\
& = \frac{45}{45+360} \approx 0.111
\end{align*}
So, normally there is only approximately a 1.369 percent chance of rain (5/365) on a given day but given that the weatherman has forecast rain, the chance of rain increases to a little more than 11 percent.