$K=\large\frac{2.303}{t}$$ \log \large\frac{a}{a-x}$
$\quad= \large\frac{2.303}{t 1/4}$$ \log \large\frac{a}{\Large\frac{3a}{4}}$
$K=\large\frac{2.303}{t1/4}$$ \log \large\frac{4}{3}$
$K= \large\frac{2.303 \times 0.125}{t 1/4}$
$\qquad= \large\frac{0.29}{t1/4}$
Hence b is the correct answer.