Explanation

(c) $n \propto \sqrt{T} \Rightarrow \frac{\mathrm{\Delta}n}{n} = \frac{\mathrm{\Delta}T}{2T}$
If tension increases by 2%, then frequency must increases by 1%.
If initial frequency $n_{1} = n$ then final frequency $n_{2} - n_{1} = 5$
$\Rightarrow \frac{101}{100}n - n = 5 \Rightarrow n = 500Hz$
Short trick : If you can remember then apply following formula to solve such type of problems.
Initial frequency of each wire $(n)$
$= \frac{(Number\ of\ beats\ heard\ per\ sec) \times 200}{(per\ centage\ change\ in\ tension\ of\ the\ wire)}$
Here $n = \frac{5 \times 200}{2} = 500Hz$