Answer
a) $\approx 50800\;\rm year$
b) $\approx 10\;\rm million \;person$
Work Step by Step
a) According to my own weight, which is 85 kg, we have about 85 kg of water and we need to find the number of protons on them.
We know that the water molecule has about 10 protons, 2 from the two hydrogen atoms and 8 from the oxygen atom. We also know that there are about $(6.02\times 10^{23})$ water molecules in one mole of water.
Knowing that the mass of one mole of water is about 18 grams.
Thus the number of water molecules in 85 kg (85000 g) of water.
$$N=\dfrac{85000}{18}\times6.02\times 10^{23}\times 10$$
$$N= \bf 2.843\times 10^{28}\;\rm protons $$
We can assume that the ray of decay is constant since the time is much less than the half-life as the new theories suggest.
Thus,
$$\dfrac{\Delta N}{\Delta t}=\lambda N$$
$$\dfrac{\Delta N}{\Delta t}=$$
Solving for $\Delta t$;
$$\Delta t=\dfrac{\Delta N}{\dfrac{\ln 2}{T_{^{\frac{1}{2}}}}N}$$
$$\Delta t=\dfrac{\Delta N}{N}\;\dfrac{T_{^{\frac{1}{2}}}}{\ln 2}$$
Plugging the known;
$$\Delta t=\dfrac{1}{2.843\times 10^{28}}\;\dfrac{10^{33}}{\ln 2}$$
$$\Delta t=\color{red}{\bf 5.0745\times 10^4}\;\rm year$$
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b) If we have 7 billion people
We know that the proton needs about 50800 years to decay and we know that the average lifetime of one person is about 70 years, so we need to find the number of people that the proton lives in to decay.
$$N=\dfrac{50800}{70}=726\;\rm person$$
Now we need to divide the 7 billion people by the 726 to find out how many people on Earth would experience the proton decay during their lifetime.
$$N_{f}=\dfrac{7\times 10^9}{726}=9.64\times 10^{6}\;\rm person$$
which is about 10 million person.
$$N_{f} \approx \color{red}{\bf10\;\rm m}\rm \;person$$
