Answer
See the detailed answer below.
Work Step by Step
We know that
$$R=\lambda N_{^{40}K}$$
solving for $N_{^{40}K}$; where $\lambda=\dfrac{\ln 2}{T_{^{\frac{1}{2}}}}$
$$N_{^{40}K}=\dfrac{RT_{^{\frac{1}{2}}}}{\ln 2}$$
Plugging the known;
$$N_{^{40}K}=\dfrac{42((1.248\times 10^9)(365.25\times 86400)}{\ln 2}$$
$$N_{^{40}K}= \bf 2.386\times 10^{18}\;\rm nuclei$$
Now we need to find the mass amount of $\rm ^{40}K$ in the one-liter of milk.
$$m_{^{40}K}=N_{^{40}K}M$$
where $M$ is the mass unit of $\rm ^{40}K$;
$$m_{^{40}K}=( 2.386\times 10^{18})(39.963998\times 1.67\times 10^{-27})$$
$$m_{^{40}K}=\color{red}{\bf 1.592\times 10^{-7}}\;\rm kg$$
Now we need to use Appendix B, as the author mentioned.
In it, we found that in a potassium sample we can see that 93.2581$\%$ is $\rm ^{39}K$ and $0.0117\%$ is $\rm ^{40}K$
Thus,
$$\dfrac{N_{^{39}K}}{N_{^{40}K}}=\dfrac{0.93 2581}{0.000117}$$
Hence,
$$ N_{^{39}K} =\dfrac{0.93 2581}{0.000117}N_{^{40}K}$$
Plugging the known;
$$ N_{^{39}K} =\dfrac{0.93 2581}{0.000117}\times 2.386\times 10^{18}$$
$$ N_{^{39}K} =\bf 1.902\times 10^{22}\;\rm nuclei$$
Now it is easy to find the mass of the mass amount of $\rm ^{39}K$ in the one-liter of milk.
$$m_{^{39}K}=N_{^{39}K}M$$
where $M$ is the mass unit of $\rm ^{39}K$;
$$m_{^{39}K}=( 1.902\times 10^{22})(38.963706\times 1.67\times 10^{-27})$$
$$m_{^{39}K}=\color{red}{\bf 1.238\times 10^{-3}}\;\rm kg$$
