Boron atom has an incomplete octet. It can accept a lone pair from $\mathrm{X}$ given in the question and can also donate electron to it, thus it can form a $\pi$-bond also known as back bonding.

The extent of back bonding depends upon the electronegativity of $\mathrm{X}$ and on the size of the valence shell. Lesser is the electronegativity of $\mathrm{X}$ and similar is the size of valence shell of $\mathrm{X}$ and $\mathrm{B}$, more will be the extend of back bonding.

$\mathrm{Cl}$ and $\mathrm{F}$ are more electronegative than $\mathrm{OMe}$ and $\mathrm{NMe}$ group, so there extent of back bonding will be least. Among ${\mathrm{BCl}}_3$ and ${\mathrm{BF}}_3, \mathrm{Cl}$ will have least tendency to form $\pi$-bond with boron, because of large size of its $3\mathrm{p}$-orbital.

Among $\mathrm{B}(\mathrm{OMe}{)}_3$ and $\mathrm{B}{\left({\mathrm{NMe}}_2\right)}_3, \mathrm{NMe}$

will have the maximum tendency to form $\pi$-bond because $\mathrm{N}$ in $\mathrm{NMe}$ is less electronegative than $\mathrm{O}$ in $\mathrm{OMe}$ group. Thus, the tendency to form $\pi$-bond with boron follows the order.

${\mathrm{BCl}}_3<{\mathrm{BF}}_3<\mathrm{B}(\mathrm{OMe}{)}_3<\mathrm{B}{\left({\mathrm{NMe}}_2\right)}_3$