A positive ion with exactly one electron ejects this electron when...

The ionization of a hydrogen-like ion corresponds to the transition of an electron from $n = 1$ to $n = \infty$.

Using the Rydberg formula:

$\dfrac{1}{\lambda} = R Z^2 \left(\dfrac{1}{1^2} - \dfrac{1}{\infty^2}\right)<br><br>\dfrac{1}{\lambda} =$R Z^2$<br><br>Substituting$\lambda = 228 \text{ \AA} = 228 \times 10^{-10} \text{ m} and the Rydberg constant $R \approx 1.097 \times 10^7 \text{ m}^{-1}$:

$Z^2 =$\dfrac{1}{\lambda R} = \dfrac{1}{228 \times 10^{-10} \times 1.097 \times 10^7}

$Z^2 =$\dfrac{1000}{250.116} \approx$4$

$Z = 2$

The ion with atomic number $Z = 2$ and exactly one electron is $\text{He}^+$.