${{\text{N}}_{\text{A}}} = {{\text{N}}_{{\text{OA}}}}{{\text{e}}^{ – \lambda {\text{t}}}} = \frac{{{{\text{N}}_{{\text{OA}}}}}}{{{2^{t/{t_{1/2}}}}}} = \frac{{{{\text{N}}_{{\text{OA}}}}}}{{{2^6}}}$

$\therefore$ Number of nuclei decayed

$ = {N_{OA}} – \frac{{{N_{OA}}}}{{{2^6}}} = \frac{{63{N_{OA}}}}{{64}}$

${{\text{N}}_{\text{B}}} = {{\text{N}}_{{\text{OB}}{{\text{e}}^{ – \lambda t}}}} = $ $\frac{{{{\text{N}}_{{\text{OB}}}}}}{{{2^{t/{t_{1/2}}}}}} = \frac{{{{\text{N}}_{{\text{OB}}}}}}{{{2^3}}}$

$\therefore$ Number of nuclei decayed

$ = {N_{OB}} – \frac{{{N_{OB}}}}{{{2^3}}} = \frac{{7{N_{OB}}}}{8}$

since, $\mathrm{N}_{\mathrm{OA}}=\mathrm{N}_{\mathrm{OB}}$

$\therefore $ Ratio of decayed numbers of nuclei

$A$ and $B = \frac{{63{N_{OA}} \times 8}}{{64 \times 7{N_{{\text{OB}}}}}} = \frac{9}{8}$