Two radioactive materials A and B have decay constants 25λ and 16λ...

According to radioactive decay formula, $N = {N}_{0}{e}^{- \lambdat}$, here, ${N}_{0}$ is the initial number of nuclei in a radioactive material.

Given here, ${\lambda}_{1} = 25\lambda$ and ${\lambda}_{2} = 16\lambda$.

And $\dfrac{{N}_{B}}{{N}_{A}} = e$

For radioactive material $A$, ${N}_{A} = {N}_{0}{e}^{- {\lambda}_{1}t}$

For radioactive material $B$, ${N}_{B} = {N}_{0}{e}^{- {\lambda}_{2}t}$

Thus, the ratio $\dfrac{{N}_{B}}{{N}_{A}} = \dfrac{{e}^{- {\lambda}_{2}t}}{{e}^{- {\lambda}_{1}t}} = {e}^{- {\lambda}_{2}t} \cdot {e}^{{\lambda}_{1}t}$

$⇒ {e}^{1} = {e}^{\left({\lambda}_{1} - {\lambda}_{2}\right)t}$

$⇒ \left({\lambda}_{1} - {\lambda}_{2}\right)t = 1$

Time, $t = \dfrac{1}{{\lambda}_{1} - {\lambda}_{2}} = \dfrac{1}{25\lambda - 16\lambda} = \dfrac{1}{9\lambda}$.

Hence, the value of $a = 9$.