The electronegativity of a carbon atom depends on its hybridisation. The correct order is

An $s$ orbital is held closer to the nucleus than a $p$ orbital. A hybrid orbital with greater $s$ character (50% in $sp$, 33% in $s{p}^2$, 25% in $s{p}^3$) holds electrons more tightly, so the carbon is more electronegative in the order $sp>s{p}^2>s{p}^3$.