25 % of the population are smokers. A smoker has 27 times more...

Given,

In a city, $25$ of the population is smoker

So, $P\left(\text{smoker}\right) = 0 . 25$

$P\left(\text{non-smoking}\right) = 0 . 75$

And smoker has $27$ times more chance of being diagnosed with lung cancer,

So, if $P\left(\dfrac{\text{Lung cancer}}{\text{non-smoker}}\right) = p$ then $P\left(\dfrac{\text{Lung cancer}}{\text{smoker}}\right) = 27p$,

Now, a person is selected at random and found to be diagnosed with lung cancer,

So, total probability of having a lung cancer will be

$P\left(\text{Lung cancer}\right) = 0 . 25 \times 27p + 0 . 75 \times p$

So the probability of him being smoker (using bayes' theorem) will be $P\left(\dfrac{\text{smoker}}{\text{Lung cancer}}\right) = \dfrac{0 . 25 \times 27p}{0 . 25 \times 27p + 0 . 75 \times p} = \dfrac{27}{30} = \dfrac{9}{10}$

Now comparing with $\dfrac{k}{10} = \dfrac{9}{10}$,

Then the value of $k$ will be $9$.