b
For \(A_{t / 2}=20\, min\), \(t=80\,min\), number of half lifes \(n=4\)

\(\therefore \) Nuclei remaining \(=\frac{\mathrm{N}_{\mathrm{o}}}{2^{4}} .\) Therefore nuclei decayed \(=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{4}}\)

For \(\mathrm{B}_{\mathrm{t} / 2}=40 \mathrm{min} ., \mathrm{t}=80 \mathrm{min},\) number of half lifes \(\mathrm{n}=2\)

\(\therefore \) Nuclei remaining \(=\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}} .\) Therefore nuclei decayed \(=\mathrm{N}_{0}-\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}}\)

\(\therefore\) Required ratio\( = \frac{{{\text{No}} - \frac{{{\text{No}}}}{{{2^4}}}}}{{{\text{No}} - \frac{{{\text{No}}}}{{{2^2}}}}} = \) \(\frac{{1 - \frac{1}{{16}}}}{{1 - \frac{1}{4}}} = \frac{{15}}{{16}} \times \frac{4}{3} = \frac{5}{4}\)