In which case effective nuclear charge is minimum?

Effective nuclear charge,${\mathrm{Z}}_{\mathrm{eff}}=\mathrm{Z}-\mathrm{S}$
where, Z- atomic number and S = screening constant
= 0.35 per electron for electron in nth orbit.
= 0.85 per electron for electrons in (n - 1)th orbit
= 1.00 per electron for electrons in (n - 2)th, (n - 3)th, (n - 4)th orbit
= 0.30 per electron in Ls-orbital (when alone)
Be, Ls² 2s² one valence electron in 2s is screened by one electron in 2s-orbital (nth orbit) and two electrons in 1s orbital ((n - 1)th orbit.)
$\begin{array}{l}\therefore \mathrm{S}=0.35+2\times 0.85=2.05\\ \therefore {\mathrm{Z}}_{\text{eff }}=4-2.05=1.95 \left(\mathrm{given}\right)\end{array}$
Be⁺ 1s² 2s¹electron is screened by two electrons in 1 s-orbital ((n - 1)th orbital) 
$\begin{array}{r}\mathrm{S}=2\times 0.85=1.70\\ {\mathrm{Z}}_{\text{eff }}=4-1.70=2.30\end{array}$
Be²⁺, 1s² one s-electron in ls-orbital (alone exists) is screened by another electron in same orbit. Thus,
$\begin{array}{l}\mathrm{S}=0.30\\ {\mathrm{Z}}_{\text{eff }}=4-0.30=3.70\end{array}$
${\mathrm{Be}}^{3+},1{\mathrm{s}}^1$no screening (single electron)
Thus, ${\mathrm{Z}}_{\mathrm{eff} }=4.$
Thus, ${\mathrm{Z}}_{\text{eff }}:\mathrm{Be}<{\mathrm{Be}}^{+}<{\mathrm{Be}}^{2+}<{\mathrm{Be}}^{3+}$