$K= \large\frac{0.693}{t_{Y2}}$
$\quad= \large\frac{0.693}{6.93}$
$\quad= minute^{-1}$
$K= \large\frac{2.303}{t} $$ \log \large\frac{100}{100-99}$
[if $a= 100, x=99\;and a-x=1]$
$\therefore \large\frac{0.693}{6.93}= \frac{2.303}{t} $$\log \large\frac{100}{1}$
$t= 46.06\;minute$
Hence C is the correct answer