To solve this problem, we need to use the formula for the energy of an electron in a hydrogen-like ion:$\\ E_n = -\frac{13.6 \, \mathrm{eV} \cdot Z^2}{n^2} \\$where $E_n$ is the energy of the electron in the $n$th energy level,$\\ Z$ is the atomic number of the ion, and $n$ is the principal quantum number.$\\$Given:$\\$• For He$^+$ ion, $Z = 2$ and the energy in the ground state $(n = 1)$ is $-x \, \mathrm{J}. \\$We need to find the energy for an electron in the $n = 2$ state for Be$^{3+}$ ion.$\\$For Be$^{3+}$ ion, $Z = 4. \\$Using the formula:$\\ E_n = -\frac{13.6 \, \mathrm{eV} \cdot Z^2}{n^2} \\$For Be$^{3+}$ ion in the $n = 2$ state:$\\ E_2 = -\frac{13.6 \, \mathrm{eV} \cdot 4^2}{2^2} \\ E_2 = -\frac{13.6 \, \mathrm{eV} \cdot 16}{4} \\ E_2 = -54.4 \, \mathrm{eV} \\$Now, let's find the energy for He$^+$ ion in the ground state:$\\ E_1 = -\frac{13.6 \, \mathrm{eV} \cdot 2^2}{1^2} \\ E_1 = -\frac{13.6 \, \mathrm{eV} \cdot 4}{1} \\ E_1 = -54.4 \, \mathrm{eV} \\$Since the energy of the He$^+$ ion in the ground state is given as $-x \, \mathrm{J}, \\$we have $-54.4 \, \mathrm{eV} = -x \, \mathrm{J}. \\$Therefore, the energy of the Be$^{3+}$ ion in the $n = 2$ state is also $-x \, \mathrm{J}. \\$Thus, the correct option is Option 3: $-X. \\$