Substance A has atomic mass number 16 and half life of 1 day....

Number of nuclei left after n half-lives is given by,

$N = {N}_{0}\left(\dfrac{1}{2}\right)\dfrac{t}{{t}_{\dfrac{1}{2}}}$.

As, ${\left({N}_{0}\right)}_{A} = \dfrac{320}{16} = 20moles$ and

${\left({N}_{0}\right)}_{B} = \dfrac{320}{32} = 10moles$

Then, ${N}_{A} = \dfrac{{\left({N}_{0}\right)}_{A}}{{\left(2\right)}^{\dfrac{2}{1}}} = \dfrac{20}{4} = 5$ and

${N}_{B} = \dfrac{{\left({N}_{0}\right)}_{B}}{{\left(2\right)}^{\dfrac{2}{0 . 5}}} = \dfrac{10}{{2}^{4}} = 0 . 625$

So total $N = 5 . 625$

and number of atoms $= 5 . 625 \times 6 . 023 \times {10}^{23}$

$= 3 . 38 \times {10}^{24}$