Which molecule has maximum bond angle amongst BF₃, BCl₃ and BBr₃?
With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for &sigma;-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can deduce that the bond angle is $\angle(\ce{X-B-X}) = 120^\circ$ for all of them.

Bent's rule won't have any effect on that.

There is also considerable &pi;-backdonation in these molecules, but that only has an effect on bond lengths, not bond angles.