Radioactive material 'A' has decay constant '8 $\lambda$'  and material 'B' has decay constant '$\lambda$'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A  will be e ?

The number of radioactive nuclei ‘N’ at any time t is given as $N\left(t\right)=N_0{e}^{-\lambda t}$

where $N_0$ is number of radioactive nuclei in the sample at some arbitrary time t = 0 and $\lambda$ is the radioactive decay constant.

$\text{ Given: }{\lambda }_A=8\lambda ,{\lambda }_B=\lambda ,N_{0A}=N_{0B}=N_0$

$\therefore \frac{N_B}{N_A}=\frac{e^{-\lambda t}}{e^{-8\lambda t}}$

$e=e^{-\lambda t}{e}^{8\lambda t}=e^{7\lambda t}\Rightarrow 1=7\lambda t\text{ or }t=\frac{1}{7\lambda }$

NOTE:-     

in original question paper the question was  "Radioactive material 'A' has decay constant '8 $\lambda$'  and material 'B' has decay constant '$\lambda$'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A  will be$\frac{1}{e}$ ? "

with this data value for  t we get after solving we get t = $-\frac{1}{7\lambda }$ 

Negative value of time is not possible. So given ratio in question should be $\frac{N_B}{N_A}=e$