How do I find the expiration time of drug in the following example?
Your general approach to the calculation looks great.  My only question is, what is the meaning of $x$ in your calculation?  

Here's how I would do the problem.

$$\ce{A-&gt;B}$$

$$A = A_0 e^{-kt}$$

$$\frac{A}{A_0} = e^{-kt}$$

$$0.35 = e^{-(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t}$$

$$ \ln{0.35} = -(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t$$ 

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})}$$

If I did the units conversion from $\mathrm{s^{-1}}$ right, then the rate constant in units of $\mathrm{{yr}^{-1}}$ is 0.316.  (I just multiplied by 60&times;60&times;24&times;365.25).

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 0.316~{\mathrm {yr}^{-1}}})}$$

When I plug in the numbers, I get 3.2 years.  It's possible I did something wrong, but my answer is way closer to yours than to the given answer.  

Here are some possible mistakes that the original writer of the question could have made to give the wrong, lower, answer.  


The question has a bunch of superfluous information about the chirality of the drug.  Mistakenly thinking that this is important to the question, perhaps somebody divided by 2 to "adjust" for chirality.  But this is not the correct approach.  We're given the activity of the drug and the decay rate of activity, so the structure of the molecule is completely irrelevant to the problem.
Another possible error is misinterpreting "is reduced to 35 % of original" mistakenly as "is reduced by 35%".  That leads to the calculation for $t$ being $\frac{-\ln{0.65}}{0.316}$ and gives the expected (but wrong) answer.


The bottom line is I think you mostly did the calculation correctly and the given answer seems incorrect, at least given the wording of the question as you supplied it.