Respective order of strength of back-bonding and Lewis acidic strength in boron trihalides is :

Respective order of strength of back-bonding and Lewis acidic strength in boron trihalides is,

${\mathrm{BF}}_3>{\mathrm{BCl}}_3>{\mathrm{BBr}}_3$ and ${\mathrm{BF}}_3<{\mathrm{BCl}}_3<{\mathrm{BBr}}_3$

Greater electronegativity of fluorine is mainly responsible for the partial positive charge on the B atom in ${\mathrm{BF}}_3,$, the fluorine atoms' capacity to donate their own lone pair to the vacant boron orbitals, generating $\rho \left(\pi \right)-p\left(\pi \right)b$ back bonding, reduces the inclination to receive a lone pair. 

Due to the effective overlap of the 2p (F) and 2p (B) orbitals, F forms the strongest back bond. As a result, the boron's electron shortage is made up for and the Lewis acidity is reduced.

However, the amount of back-bonding reduces as orbital size grows. This is a result of the inefficient overlap between the 2p orbital of B and the 3p and 4p orbitals of $\mathrm{Cl}$ and $\mathrm{Br}$, respectively.

Hence, option C is correct.