In order to estimate f. the true fraction of smokers in a large population. Someone selects n people at random. His estimator Mn is obtained by dividing Sn. The number of smokers in his sample by n, i.e., Mn = Sn/n. So, he chooses the sample size n to be the smallest number for which the Chebyshev inequality yields a guarantee that:
$$P(|M_n-f|\ge\varepsilon)\le\delta$$
Where epsilon and delta are some prespecified tolerances. Determine how the value of n recommended by the Chebyshev inequality changes in the following cases.
a) The value of epsilon is reduced to half its original value. b) The probability delta is reduced to half its original value.
As discussed in comments, the random variable $M_n$ has mean $f$ and standard deviation $\sigma_n=\frac{\sqrt{f(1-f)}}{\sqrt{n}}$.
The standard form of the Chebyshev Inequality (please see Wikipedia) says that $$\Pr(|M_n-f|\ge k\sigma_n)\le \frac{1}{k^2}.\tag{1}$$ Suppose that we want to leave $\delta$ unchanged, but want to shrink $\epsilon$ by a factor of $2$. So in the notation of (1) we want to leave $\frac{1}{k^2}$ unchanged, meaning we want $k$ to be unchanged.
Since the original $\epsilon$ is $k\frac{\sqrt{f(1-f)}}{\sqrt{n}}$, to shrink it by a factor of $2$ while leaving $k$ unchanged, we need to multiply $\sqrt{n}$ by $2$, so we need to multiply $n$ by $4$. Shrinking the width of the confidence interval is expensive!
Now it's your turn to find how $n$ should change if $\epsilon$ is unchanged, but $\delta$ changes to half its previous value.