Correct option is A. $$9:8$$
$$N_A = N_{OA} e^{-\lambda t} = \dfrac{N_{OA}}{2^{t/t}\dfrac{1}{2}} = \dfrac{N_{OA}}{2^6}$$
$$\therefore$$ Number of nuclei decayed
$$= N_{OA} - \dfrac{N_{OA}}{2^6} = \dfrac{63N_{OA}}{64}$$
$$N_B = N_{OB}e^{-\lambda t} = \dfrac{N_{OB}}{2^{t/t} \dfrac{1}{2}} = \dfrac{N_{OB}}{2^3}$$
$$\therefore$$ Number of nuclei decayed
$$= N_{OB} - \dfrac{N_{OB}}{2^3} = \dfrac{7N_{OB}}{8}$$
Since $$N_{OA} = N_{OB}$$
$$\therefore$$ ratio of decayed numbers of nuclei
$$A \& B = \dfrac{63 N_{OA} \times 8}{64 \times 7N_{OB}} = \dfrac{9}{8}$$