Step-by-step Solution:

1. Understand the Core Principle: The problem involves two mirror nuclei, $_7^{15}N$ and $_8^{15}O$. These nuclei have the same mass number (A=15), and the number of protons in one is equal to the number of neutrons in the other, and vice versa. It is assumed that the strong nuclear force, which is charge-independent, contributes equally to the binding energy of both nuclei. Therefore, the difference in their binding energies arises solely from the difference in their electrostatic (Coulomb) repulsion energy. The nucleus with more protons ($^{15}O$) will have a greater electrostatic repulsion, which in turn reduces its binding energy. Thus, we can state the relationship: $\Delta BE = BE(^{15}N) - BE(^{15}O) = E_{elec}(^{15}O) - E_{elec}(^{15}N) = \Delta E_{elec}$

2. Calculate the Difference in Binding Energy ($\Delta BE$): The binding energy (BE) of a nucleus can be calculated from the given atomic masses. For a nucleus $^A_ZX$, the binding energy is: $BE = [Z \cdot m(_1^1H) + (A-Z) \cdot m_n - M(^A_ZX)] c^2$ where $m(_1^1H)$ is the mass of a hydrogen atom and $m_n$ is the mass of a neutron.

   For $_7^{15}N$ (Z=7, A=15): $BE_N = [7 \cdot m(_1^1H) + 8 \cdot m_n - M(^{15}N)] c^2$ For $_8^{15}O$ (Z=8, A=15): $BE_O = [8 \cdot m(_1^1H) + 7 \cdot m_n - M(^{15}O)] c^2$

   The difference in binding energy is: $\Delta BE = BE_N - BE_O = [(-m(_1^1H) + m_n) - (M(^{15}N) - M(^{15}O))]c^2$ $\Delta BE = [ (M(^{15}O) - M(^{15}N)) + (m_n - m(_1^1H)) ]c^2$

   Substitute the given mass values (in atomic mass units, u):

   • $M(^{15}O) = 15.003065\text{ u}$
   • $M(^{15}N) = 15.000109\text{ u}$
   • $m_n = 1.008665\text{ u}$
   • $m(_1^1H) = 1.007825\text{ u}$

   $\Delta BE = [ (15.003065 - 15.000109) + (1.008665 - 1.007825) ] \text{ u} \cdot c^2$ $\Delta BE = [ 0.002956 + 0.000840 ] \text{ u} \cdot c^2$ $\Delta BE = 0.003796 \text{ u} \cdot c^2$ Using the conversion factor 1 u = 931.5 MeV/c²: $\Delta BE = 0.003796 \times 931.5 \text{ MeV} \approx 3.5357 \text{ MeV}$

3. Calculate the Difference in Electrostatic Energy ($\Delta E_{elec}$): The electrostatic energy of a nucleus is given by the formula: $E_{elec} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi {\varepsilon _0}R}}$ For $_7^{15}N$ (Z=7): $E_{elec}(N) = {3 \over 5}{{7(7 - 1)} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right) = {3 \over 5}{{42} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right)$ For $_8^{15}O$ (Z=8): $E_{elec}(O) = {3 \over 5}{{8(8 - 1)} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right) = {3 \over 5}{{56} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right)$

   The difference in electrostatic energy is: $\Delta E_{elec} = E_{elec}(O) - E_{elec}(N) = {3 \over 5} {{(56 - 42)} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right)$ $\Delta E_{elec} = {3 \over 5} {{14} \over R} \left( {{e^2} \over {4\pi {\varepsilon _0}}} \right)$ Given that ${{e^2} / ({4\pi {\varepsilon _0}})} = 1.44 \text{ MeV fm}$: $\Delta E_{elec} = {3 \over 5} \times {{14} \over R} \times 1.44 \text{ MeV fm}$ $\Delta E_{elec} = {{42 \times 1.44} \over {5R}} \text{ MeV fm} = {{60.48} \over {5R}} \text{ MeV fm} = {{12.096} \over R} \text{ MeV fm}$

4. Solve for the Radius (R): Now, we equate the two differences: $\Delta BE = \Delta E_{elec}$ $3.5357 \text{ MeV} = {{12.096} \over R} \text{ MeV fm}$ Solving for R: $R = {{12.096} \over {3.5357}} \text{ fm}$ $R \approx 3.421 \text{ fm}$

5. Conclusion: The calculated radius is approximately 3.42 fm. This corresponds to option C.