Answer
(a) $\,^{222}_{86}Rn\rightarrow \,^{218}_{84}Po+\,^{4}_{2}\alpha$
(b) 8.87 days
Work Step by Step
(a) Alpha particle has mass=4 and charge=2.
When $\alpha$ particle is emitted, mass of the nuclide is decreased by 4 and charge by 2. So Rn-222 decays to the nuclide with mass number=(222-4)=218 and atomic number=(86-2)=84. The nuclide formed is $\,^{218}_{84}Po$. The balanced nuclear equation is
$\,^{222}_{86}Rn\rightarrow \,^{218}_{84}Po+\,^{4}_{2}\alpha$
(b) Recall that $\ln(\frac{A_{0}}{A})=kt=\frac{0.693}{t_{1/2}}\times t$ where $A_{0}$ is the initial activity and $A$ is the activity after time $t$.
$\implies \ln(\frac{100}{20})=1.60944=\frac{0.693}{3.82\,d}\times(t)$
Or $t=\frac{1.60944\times3.82\,d}{0.693}=8.87\,d$
