The correct option is C (iv)<(ii)<(iii)<(i)
Basicity is inversely proportional to electronegativity. Here CN− is most stable anion and hence least basic because negative charge is present on highly electronegative N atom.
Negative charge in (i) is most destabilised because of electron donating (methyl) groups on C. So it is most basic in nature.
Comparing (ii) and (iii), as the % s-character in hybridisation of C is more, more is the electronegativity, more is the stability of anion and hence lesser is the basicity. Hybridisation of C in alkyne and alkane is sp and sp3 respectively. So negative charge on alkyne; with maximum s−character, its anion is more stabilised and is less basic than anion of alkane for the same reason.
Hence, (c) is the correct option.