A He^+ ion (hydrogen-like, Z=2) in its n=4 state makes a transition directly to the n=2 state and emits one photon. This photon falls on a metal…

The energy levels of a hydrogen-like ion are $E_n=-13.6Z^2/n^2$ eV. For $\mathrm{He^+}$ ($Z=2$), $E_4=-13.6\times4/16=-3.4$ eV and $E_2=-13.6\times4/4=-13.6$ eV. The emitted photon therefore has energy $h\nu=E_4-E_2=10.2$ eV. A common error is to use the hydrogen value (2.55 eV) without the $Z^2$ factor, which would give a photon energy below the work function and no emission at all.

By Einstein's photoelectric equation, $K_{max}=h\nu-\phi=10.2-4.2=6.0$ eV. The stopping potential satisfies $eV_0=K_{max}$, so $V_0=6.0$ V. The answer is $6$.