We assume independence, which is **very** unreasonable. But without some assumption we cannot solve the problem.
The random variable $X$ is the number of consecutive rainy days after a rainless day. So $X$ takes on values $0,1,2,3,\dots$.
The probability that $X=0$ is $0.6$.
For $X=1$, we need RN (rain, then not rain). The probability of this is $(0.4)(0.6)$.
For $X=2$, we need RRN. The probability of this is $(0.4)^2(0.6)$.
And so on.