Problem on Chebychev's inequality. 
I computed $\sigma=.05$
i got $n=.001$ 
i want to know whats going on in this question.I am not having any idea where to start where to end. 

According to Chebychev's Inequlaity for the said problem
$\mu = \frac{n}{2}$
$\sigma = \frac{\sqrt{n}}{2}$
Thus
$$P({|S_n - \frac{n}{2}}| \ge 2. \frac{\sqrt{n}}{2}) \le 1-\frac{3}{4} = \frac{1}{2^2}$$
$$P(\frac{|2S_n - n|}{2}| \ge 2. \frac{\sqrt{n}}{2}) \le 1-\frac{3}{4} = \frac{1}{2^2}\tag 1$$
In the given expression
You have 
$$P(\frac{|2S_n - n|}{2n}| \ge .1) \le 1-\frac{3}{4} = \frac{1}{2^2}$$
$$P(\frac{|2S_n - n|}{2}| \ge .1n) \le 1-\frac{3}{4} = \frac{1}{2^2}\tag 2$$
Compare (1) and (2)
and equate $$0.1n = \sqrt{n}\implies n = 100$$
