Probability of an ectopic pregnancy An ectopic pregnancy is twice as likely to occur when the pregnant woman is a smoker than when she is a nonsmoker. If 32% of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers?
Attempt: Given the women has had an ectopic pregnancy, we can condition on her being a smoker. So I believe the prob we want in this question is P(woman smoke| she had ectopic preg).
Denote ectopic preg by EP
This is equal to P(EP and woman smoke)/P(EP), which can be rewritten as P(EP|woman smoke)/(P(woman smoke)P(EP)) by Bayes.
Then taking into consideration data in the question, I said the above can further be written as 2P(EP|woman not smoke)/(P(woman smoke)P(EP)).           (1)
Using the total Prob law on P(EP) gives P(EP) = P(EP|woman smoke)P(woman smoke) + P(EP|woman did not smoke)P(woman did not smoke), which is equal to P(EP|woman did not smoke)[2P(woman smoke) + P(woman did not smoke)]
Subbing this into (1)  and cancelling gives 2/(.32)(0.64 + .68) which is > 1.
There is probably a simpler method to mine, but I also want to know where I went wrong.
Many thanks. 

Let $E$ denote the event of an ectopic pregnancy and let $S$ denote the event that the woman is a smoker. So far we are given
$$\Pr(E\mid S) = 2\Pr(E\mid\overline{S}),\ \ \ \Pr(S)=0.32$$
We wish to find $P(S\mid E)$. From Bayes' law
$$\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)\Pr(S)}{\Pr(E\mid S)\Pr(S) + \Pr(E\mid \overline{S})\Pr(\overline{S})}$$
Making a few substitutions, we have
$$\frac{\Pr(E\mid S)\Pr(S)}{\Pr(E\mid S)\Pr(S) + 0.5\Pr(E\mid S)\left(1-\Pr(S)\right)}=\frac{2\Pr(S)}{1+\Pr(S)}=\frac{0.64}{1.32}$$
The final probability comes down to
$$\Pr(S\mid E) = \frac{16}{33}\approx 0.485$$
Where you went wrong seems to be in applying Bayes' law. You made the substitution
$$\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)}{\Pr(E)\color{red}{\Pr(S)}}$$
The term $\Pr(S)$ is on the numeartor, not the denominator. It should be
$$\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)\color{green}{\Pr(S)}}{\Pr(E)}$$
