Explanation

(c)
Given ,$N_{0}$=initial concentration =1
N=concentration after time $t = \frac{1}{8}$
K=$6.9 \times 10^{- 3}s^{- 1}$
$N = N_{0}\left( \frac{1}{2} \right)^{n}$
Or
$\frac{1}{8} = 1\left( \frac{1}{2} \right)^{n}$
$\therefore n = 3$
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{6.9 \times 10^{- 3}} = 100s$
T=$n \times t_{1/2}$
=3$\times 100$
=300s
∴ after 300s it will be reduced to $1/8$ of original concentration.