= Solution
{c}
Put $k_Z=Ze^2/(4\pi\epsilon_0)$. <Gauss's law> for a uniform spherical <nuclear charge distribution>, with the <electrostatic potential> matched at $R$, gives the electronic <potential energy>
$$
V_{\rm sph}(r)=\begin{cases}
-\dfrac{k_Z}{2R}(3-r^2/R^2),&r<R,\\
-k_Z/r,&r\geq R.
\end{cases}
$$
Subtracting the point-nucleus <Coulomb potential> leaves the nonnegative perturbation
$$
\delta V(r)=k_Z\left(\frac1r-\frac3{2R}+\frac{r^2}{2R^3}\right)\quad(r<R),\qquad
\delta V=0\quad(r\geq R).
$$
The normalized nonrelativistic <hydrogen-like ion> $1s$ <wavefunction> has $|\psi(r)|^2=Z^3e^{-2Zr/a_0}/(\pi a_0^3)$. Since $ZR/a_0\ll1$, approximate it inside the <atomic nucleus> by $|\psi(0)|^2$. First-order <time-independent perturbation theory> then gives
$$
\begin{aligned}
\Delta E_{1s}&\simeq4\pi|\psi(0)|^2 k_Z\int_0^R\left(r-\frac{3r^2}{2R}+\frac{r^4}{2R^3}\right)dr\\
&=4\pi|\psi(0)|^2 k_Z\frac{R^2}{10}
=\frac25\frac{Z^4e^2}{4\pi\epsilon_0a_0}\left(\frac R{a_0}\right)^2.
\end{aligned}
$$
Define $E_H=-Z^2\mathrm{Ry}$, as the paper specifies for the <hydrogen-like ion>, rather than the neutral-<hydrogen> value. Using $\mathrm{Ry}=e^2/(8\pi\epsilon_0a_0)$ gives
$$
\boxed{\Delta E_{1s}=\frac45Z^2|E_H|\left(\frac R{a_0}\right)^2
=\frac45 Z^4\mathrm{Ry}\left(\frac R{a_0}\right)^2}
$$
to leading order in $ZR/a_0$. For $A=208$, $R\simeq7.11\,\mathrm{fm}$, $ZR/a_0\simeq0.0110$, and
$$
\boxed{\Delta E_{1s}\simeq+8.88\,\mathrm{eV},\qquad
\Delta E_{1s}/h\simeq+2.15\times10^{15}\,\mathrm{Hz}.}
$$
The finite <nuclear charge radius> makes the <energy level> less bound. The quoted <frequency> is a level shift divided by $h$: a line frequency shifts by $(\Delta E_u-\Delta E_l)/h$, so a decay into this shifted $1s$ level alone has the opposite sign.
The formula is a nonrelativistic, small-<nuclear charge radius> estimate. Keeping the exponential in the first-order integral gives $\Delta E_{1s}=\Delta E_{\rm leading}[1-5ZR/(6a_0)+O((ZR/a_0)^2)]$, about a one-percent correction here. More seriously, $Z\alpha_{\rm fs}\simeq0.598$ for <lead>, so relativistic contraction is substantial. A precise physical prediction needs a finite-nucleus <Dirac equation> calculation; the boxed estimate is the model requested, not precision heavy-ion spectroscopy.
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