$K_2= \large\frac{2.303}{t_2}$$ \log \large\frac{100}{50}$ for $50 \%\;B $ reacted.

$K_1= \large\frac{2.303}{t_1}$$ \log \large\frac{100}{6}$ for $94 \%\;A $ reacted.

$\therefore \large\frac{K_2}{K_1} =\frac{t_1}{t_2} \times \frac{0.3010}{1.2218}$

Since $t_2=t_1$ because $50\%\;B$ has reacted when $94 \%$ A has reacted.

$\therefore \large\frac{K_2}{K_1}=\frac{0.3010}{1.2218}$$=0.246$

and $\large\frac{K_1}{K_2}$$=4.006$

Hence c is the correct answer.