Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to calculate the Time to Boil the Water. So we need to find the Energy Released Per Decay.
The decay of $ ^{223}\text{Ra} $ releases energy when it emits an alpha particle and transforms into $ ^{219}\text{Rn} $.
We know that
$$E_{\rm released}=\Delta mc^2$$
where $ c^2 = 931.49 \, \text{MeV}/\text{u} $, and the mass difference is given by $\Delta m=m_{( ^{223}\text{Ra})} - m_{(^{219}\text{Rn})} - m_{(^4\text{He}) }$;
$$
E_{\rm released} = \left[ m_{( ^{223}\text{Ra})} - m_{(^{219}\text{Rn})} - m_{(^4\text{He}) }\right] (931.49 )
$$
Substituting the known:
$$
E_{\rm released} = [223.018499 - 219.009477 - 4.002602 ](931.49 )
$$
$$
E_{\rm released} =\bf 5.98 \; \rm {MeV} =\bf 9.581 \times 10^{-13} \rm \; J\tag 1
$$
This means that each alpha decay process, it releases $ 9.581 \times 10^{-13} \, \text{J} $.
Now we need to calculate the Energy Needed to Heat the 100 mL (or 0.1 kg where $m=\rho /V$) of water from 18$^\circ$C to 100$^\circ$C.
So,
$$ Q = mc\Delta T $$
Substituting the known:
$$
Q = (0.1 ) (4190 ) (100 - 18) =\bf 34400 \rm \, {J}\tag 2
$$
Now we need to calculate the Number of Decays Needed to provide a $ 34,400 \, \text{J} $ of energy.
$$
N_\text{decays} = \frac{Q}{E_{\rm released}} $$
Plug from (1) and (2);
$$
N_\text{decay} = \frac{34400 }{9.581 \times 10^{-13} }=\bf 3.59 \times 10^{16}\;\rm decay
$$
Now we need to find the Initial Number of $ ^{223}\text{Ra} $ atoms.
We know that the molar mass of $ ^{223}\text{Ra} $ is 223 g/mol, so for 1.0 g:
$$
N_0 =\dfrac{m_{\rm sample}}{M}N_A$$
where $N_A$ is Avogadro's number.
Plug the known;
$$N_0= \frac{1.0 \, \text{g}}{223 \, \text{g/mol}} ( 6.022 \times 10^{23} \, \text{atoms/mol})
$$
$$
N_0 =\bf 2.70 \times 10^{21} \, \rm {atoms}
$$
Now we need to find the Time Required for $ 3.59 \times 10^{16} $ Decays, where the decay rate is given by:
$$
\frac{dN}{dt} = -rN
$$
where $ r = \dfrac{\ln(2)}{t_{1/2}} $
$$\frac{dN}{dt} = -\dfrac{\ln(2)}{t_{1/2}}N $$
Replacing $dN/dt$ by $\Delta N/\Delta t$, and solving for $\Delta t$
$$
\Delta t= \frac{\Delta N}{ -\dfrac{\ln(2)}{t_{1/2}}N }
$$
$$
\Delta t=\dfrac{-t_{1/2}}{\ln(2)} \frac{\Delta N}{ N }
$$
where $N=N_0$, and $\Delta N=- N_{\rm decay}$
$$
\Delta t=\dfrac{-t_{1/2}}{\ln(2)} \frac{ N_{\rm decay}}{ N_0 }
$$
Plug the known;
$$
\Delta t=\dfrac{-(11.43\times 24\times 3600)}{\ln(2)}\cdot \frac{-(3.59 \times 10^{16})}{( 2.70 \times 10^{21}) }
$$
$$
\Delta t= \color{red}{\bf 18.94}\;\rm s
$$
$$\color{blue}{\bf [b]}$$
Alpha particles can ionize water molecules, potentially splitting $ \text{H}_2\text{O} $ into $ \text{H}_2 $ and $ \text{O}_2 $ gases, causing bubbles. However, the water's chemistry remains largely unchanged.
