b
Let initially there are total \(\mathrm{N}_{0}\) number of nuclei

At time \(\mathrm{t} \frac{N_{B}}{N_{A}}=0.3(\text { given })\)

\(\Rightarrow \quad N_{B}=0.3 N_{A}\)

\(\mathrm{N}_{0}=N_{A}+N_{B}=N_{A}+0.3 N_{A}\)

\(\therefore \quad N_{A}=\frac{\mathrm{N}_{0}}{1.3}\)

As we know \(N_{t}=N_{0} e^{-\lambda t}\)

or, \(\frac{\mathrm{N}_{0}}{1.3}=N_{0} e^{-\lambda t}\)

\(\frac{1}{1.3}=e^{-\lambda t} \Rightarrow \ln (1.3)=\lambda t\)

or, \(t=\frac{\ln (1.3)}{\lambda} \Rightarrow t=\frac{\ln (1.3)}{\frac{\ln (2)}{T}}=\frac{\ln (1.3)}{\ln (2)} T\)