The correct order regarding the electronegativity of hybrid orbitals of carbon is:

Electronegativity of a hybrid orbital is directly proportional to its s-character. Since s-orbitals are closer to the nucleus, greater s-character results in stronger electron pull. The s-character is $50\%$ for $sp$, $33.3\%$ for $s{p}^2$, and $25\%$ for $s{p}^3$. Thus, the order is $sp>s{p}^2>s{p}^3$.