The energy levels of an electron in a hydrogen-like ion (an atom or ion with only one electron) can be quantified using the formula:

$$ E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2} $$

where:

• $ E_n $ is the energy of the electron in the nth energy level,
• $ Z $ is the atomic number (number of protons) of the ion,
• $ 13.6 \text{ eV} $ is the ionization energy of hydrogen,
• $ n $ is the principal quantum number (the energy level).

Since we need to compare this across different ions in different energy states, let's plug in some numbers:

For $ \mathrm{He}^{+} $ (Helium ion):

• $ Z = 2 $ (as helium has 2 protons)
• $ n = 1 $ for ground state

$$ E_1 = -\frac{(2)^2 \cdot 13.6 \text{ eV}}{1^2} = -4 \cdot 13.6 \text{ eV} = -54.4 \text{ eV} $$

However, the problem gives the energy in joules, and it's given a constant $ x $. So, $ x = 54.4 \text{ eV} $ (converted to joules as needed).

Next, for the $ \mathrm{Be}^{3+} $ ion:

• $ Z = 4 $ (as beryllium has 4 protons)
• $ n = 2 $

$$ E_2 = -\frac{(4)^2 \cdot 13.6 \text{ eV}}{2^2} = -\frac{16 \cdot 13.6 \text{ eV}}{4} = -54.4 \text{ eV} $$

Since initially $ x = 54.4 \text{ eV} $ (or its equivalent in joules) for $$ \mathrm{He}^{+} \text{ at } n=1 $$, and now we have the same energy for $$ \mathrm{Be}^{3+} \text{ at } n=2 $$, the energy level in joules would also equate to $ x $, just at a different energy state and ion.

So, the answer is:

Option A: $ -x $.