Step 1: Compare the given ionization energy with hydrogen's known value.
For any hydrogen-like (single electron) species, the ionization energy scales with the square of the atomic number $Z$, because a hydrogen-like ion is just a one-electron atom with nuclear charge $Z$ instead of $1$:
\[ E_Z = Z^2 \times E_H, \quad E_H = 13.6 \, \text{eV} \]
Step 2: Set up the ratio instead of rearranging the formula.
Rather than solving the equation algebraically, treat it as a straight ratio to the hydrogen value:
\[ Z^2 = \frac{E_Z}{E_H} = \frac{217.6}{13.6} \]
Step 3: Work out the ratio.
\[ Z^2 = 16 \implies Z = \sqrt{16} = 4 \]
A quick check: $Z=4$ gives $16 \times 13.6 = 217.6$ eV, which matches the given value exactly, so the ratio method holds up.
Step 4: Locate this atomic number on the periodic table.
$Z = 4$ is beryllium. Since the species behaves as hydrogen-like, three of its four electrons must already be removed, so the ion is $Be^{3+}$.
Step 5: Bring in the mass number from the stable isotope.
Beryllium occurs in nature almost entirely as a single stable isotope, $^9Be$, with mass number $A = 9$. The neutron count follows from the basic nuclear rule:
\[ \text{Neutrons} = A - Z = 9 - 4 = 5 \]
Step 6: Rule out the other options.
Options (2), (3), and (4) would only be correct for atomic numbers that do not satisfy $Z^2 = 217.6/13.6$, so they cannot be right once $Z=4$ is fixed.
Final Answer:
\[ \boxed{5 \text{ neutrons}} \]