The correct option is
B II > I > III
Conjugate base of I, II and III:
I.
HC≡C⊖(Negative charge onsp C atom) sp carbons have 50% s character. More the s- character, more will be the electronegativity. Hence it will stabilise the negative charge on carbon.
(II)
In this conjugate base, it has
6 π electrons which will obey the 4n+2 rule. Thus it is aromatic in nature. Hence it stabilise the conjugate base more than others.
(III)
In this conjugate base, it has
4 π electrons which will obey 4n
π electron rule. Hence it is anti-aromatic and it destabilise the conjugate base more.
Stability of conjugate base
∝ acidic character
Hence, the decreasing acidic character follows II > I > III