Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $sp ^2$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.

But diborane, $B _2 H _6$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.

Hence, each Boron atom is $sp ^3$ – hybridized.