Question

Ga atom loose electron(s) to form $Ga^{+}$, $Ga^{2+}$, $Ga^{3+}$ ions.Which step will have highest ionization energy?

Answer 1

"Gallium atom loose electrons to form first ionization energy  

$Ga\quad \longrightarrow \quad Ga^{ + }\quad +\quad e^{ - }$

The "second ionization energy" is the energy required to eliminate a "second electron" to form "2+ cations" from "1+ cations".

$Ga^{ + }\quad \longrightarrow \quad Ga^{ 2+ }\quad +\quad e^{ - }$

The third ionization energy is the "energy required" to form 3+ cations.

$Ga^{ 2+ }\quad \longrightarrow \quad Ga^{ 3+ }\quad +\quad e^{ - }$

The "second ionization energy" is larger than the first ionization energy, since it requires more energy to eliminate an "electron" from a cation than it is from a "neutral atom". Likewise, the "third ionization energy" is "larger" than the "second ionization energy". The $Ga^{2 + }$ has the highest ionization energy"."

Answer 2

Hey mate.....☺☺
$\huge\red{Ga^2+}$

i.e. the SECOND step have heighest IONISATION ENERGY✔

I HOPE IT IS HELPFUL☺☺