= Solution
The mirror nuclei have interchanged <neutron> and <proton> numbers. Neglect their strong-interaction binding difference and describe them with the same uniformly charged sharp sphere of <radius> $R$. The <nuclear Coulomb energy> is
$$
E_C(Z)=\frac35\frac{Z^2e^2}{4\pi\epsilon_0R}
=\frac35\frac{Z^2\alpha\hbar c}{R}.
$$
Changing one <neutron> into a <proton> decreases the nucleon rest-mass sum by $m_n-m_p=1.294\ \mathrm{MeV}/c^2$, but raises the Coulomb <energy>. The diagram gives $(M_F-M_O)c^2=2.25\ \mathrm{MeV}$ for nuclear masses, so
$$
2.25=-1.294+\frac35(9^2-8^2)\frac{\alpha\hbar c}{R}.
$$
With the standard electromagnetic conversion $\alpha\hbar c\simeq1.44\ \mathrm{MeV\,fm}$, the <uniform-sphere Coulomb displacement energy> estimate is
$$
\boxed{R=\frac{(3/5)(17)(1.44)}{3.544}\ \mathrm{fm}\simeq4.14\ \mathrm{fm}.}
$$
This $R$ is the uniform sphere's outer <radius>. If “charge <radius>” means the rms <nuclear charge radius>, the same distribution gives
$$
\boxed{\sqrt{\langle r^2\rangle}=\sqrt{3/5}\,R\simeq3.21\ \mathrm{fm}.}
$$
The numerical result is an estimate within the charge-density model, not a model-independent <radius> measurement. A discrete-proton correction replaces $Z^2$ by $Z(Z-1)$, giving $\Delta[Z(Z-1)]=16$ and $R\simeq3.90\ \mathrm{fm}$. The main answer above follows the literal continuum charged-sphere model. Strong isospin-breaking, nonuniform density and changes in the valence-nucleon distribution also limit the extraction.
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