electronegativity depends on \(s\) - character, more is the \(s\) - character, more is the electronegativity. So, the order for electronegativity should be \(sp \,> \,sp ^2\,>\,sp ^3\)

bond angle between same hybrid orbitals is in the following order

\(sp \left(180^{\circ}\right)\,>\, sp ^2\left(120^{\circ}\right)\,>\,sp ^3\left(109.5^{\circ}\right)\)

Size \(\rightarrow\) more \(s\)-character \(\rightarrow\) more compact \(\rightarrow\) smaller size.

So, the given order is correct, \(sp \,<\, sp ^2\,<\,sp ^3\).

Energy level \(\rightarrow\) since \(s\)-orbitals have lower energy closer than \(p\) - as they allow closer proximity of the \(e^{-}\)to the nuclear. Thus, more \(p\) - character more energy.

So, the order will be \(sp \,<\,sp ^2\,<\,sp ^3\)