b
The electron distribution in benzene is symmetrical and a pair of electrons  become localised at the requirement of the electrophilic reagent In biphenyl, one ring behaves as a \(+R\) group, thereby activating the other ring in the \(o-\) and \(p-\) positions. Hence, on the charge distribution approach, biphenyl can be expected to be more reactive than benzene towards electrophiles. 
The \(+R\) effect in biphenyl is transmitted from one ring to the other through  the intramolecular bond. Maximum transmission is obtained if the two rings are  coplannar, but because of the steric effects (weak) of the hydrogens in the  \(o-\) positions, the two rings are being hindered from becoming coplanar. At  the same time, biphenyl molecules in solution are subject to thermal agitation  ('bombardment' by other molecules), and this also tends to prevent the  biphenyl molecule from becoming coplanar (in which state the molecule will  have a minimum entropy). The overall result is that the resonance effect is partially nullified, i. e., the activation of one ring by the other through  resonance is relatively weak. 
If we now consider fluorene, its unsaturated system is the same as that of  biphenyl, but in the former, because it is a flat molecule due to the presence of  the middle ring (formed by the methylene bridge), the \(+R\) effect of one  benzene ring is completely transmitted to the other ring. Hence, fluorene can  be expected to be more reactive than biphenyl towards electrophiles.  
If we consider the problem from the carbonium - ion stability approach, then the more extended the conjugation and the more nearly coplanar the resonating structures, the more stable are the resonance hybrids. We therefore arrive at the same order of reactivity as before by using the same arguments.