Conditional probability question involving Bayes' theorem

Straightforward application of Bayes' Theorem. I agree with @msm that you have the right method and answer. Notice that your answer (approx 0.1) is substantially greater than just the probability (approx .003) of choosing the 'pure blue' button without conditional info from 'testing out' the button with 5 tosses.

Just for verification, I simulated the experiment 10 million times in R statistical software. Results are in reasonable agreement with your answer. (Because choosing a 'pure blue' button is so rare, such a simulated result may be accurate to only about 2, maybe 3, decimal places.)

 m = 10^7;  p = x = numeric(m)
 b = c(1, rep(.5, 299))     # buttons in jar
 for (i in 1:m) {
   p[i] = sample(b, 1)         # choose a button
   x[i] = rbinom(1, 5, p[i]) } # toss 5 times and count 'blue's
 mean(p == 1)                  # proportion of 'pure blue' buttons chosen
 ## 0.0033274
 1/300                    # compare
 ## 0.003333333
 mean(p[x==5] == 1)       # when get 5 'blues' what proportion 'pure blue' bottons
 ## 0.09672815
 1/(299/32 + 1)           # compare (your answer)
 ## 0.09667674 

It is correct. Maybe it is more clear when it is written as

$$\begin{align}P(\text{chose blue }|\text{5 blue tosses}) &= \frac{P(\text{5 blue tosses|blue})P(\text{blue})}{P(\text{5 blue tosses|blue })P(\text{blue })+P(\text{5 blue tosses|other})P(\text{other})} \\[10pt] &= \frac{(1^5)\frac{1}{300}}{(1^5)\frac{1}{300}+(\frac{1}{2})^5\frac{299}{300}}\end{align}$$