The ionisation constant of $\mathrm{NH_4^+}$ in water is $5.6 \times 10^{-10}$ at $25^{\circ}\mathrm{C}$. The rate constant for the reaction of $\mathrm{NH_4^+}$ and $\mathrm{OH^-}$ to form $\mathrm{NH_3}$ and $\mathrm{H_2O}$ at $25^{\circ}\mathrm{C}$ is $3.4 \times 10^{10} \, \mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{s}^{-1}$. The rate constant for proton transfer from water to $\mathrm{NH_3}$ is

For the equilibrium $$\mathrm{NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+}$$ the acid ionisation constant is $$K_a=5.6\times 10^{-10}$$ Consider the proton-transfer reaction with hydroxide: $$\mathrm{NH_4^+ + OH^- \rightarrow NH_3 + H_2O}$$ Its rate constant is $$k_1=3.4\times 10^{10}\,\mathrm{L\,mol^{-1}\,s^{-1}}$$ The reverse reaction is proton transfer from water to ammonia: $$\mathrm{NH_3 + H_2O \rightarrow NH_4^+ + OH^-}$$ Let its rate constant be $k_{-1}$. For this conjugate acid-base pair, $$K_b=\frac{K_w}{K_a}$$ $$K_b=\frac{10^{-14}}{5.6\times 10^{-10}}=1.786\times 10^{-5}$$ Also, for the reaction pair, $$K_b=\frac{k_1}{k_{-1}}$$ Therefore, $$k_{-1}=\frac{k_1}{K_b}$$ $$k_{-1}=\frac{3.4\times 10^{10}}{1.786\times 10^{-5}}\approx 1.90\times 10^{15}$$ This is a first-order rate constant with units $\mathrm{s^{-1}}$, but this value does not match the options, so check the equilibrium relation carefully. Using the acid dissociation directly for $$\mathrm{NH_4^+ \rightleftharpoons NH_3 + H^+}$$ we have $$K_a=\frac{k_\mathrm{d}}{k_\mathrm{r}}$$ Here the forward step corresponds to proton removal by $\mathrm{OH^-}$, and the reverse protonation of $\mathrm{NH_3}$ by water gives $$K_a=\frac{k_1K_w^{-1}}{k_{-1}}$$ Hence, $$k_{-1}=\frac{k_1K_w}{K_a}$$ $$k_{-1}=\frac{3.4\times 10^{10}\times 10^{-14}}{5.6\times 10^{-10}}=6.07\times 10^{5}\,\mathrm{s^{-1}}$$ Comparing with the given options, this matches option $A$.