The ionization constant of $NH_4OH$ is $1.77 \times 10^{-5}$ at $298\;K$ . Hydrolysis constant of it is

Question

$\begin{array}{1 1} (a)\;6.5 \times 10^{-12} \\(b)\;5.65 \times 10^{-23} \\ (c)\;5.65 \times 10^{-12} \\ (d)\;1.506 \times 10^{-20} \end{array} $