(B) False.
To construct a triangle similar to $\triangle ABC$ with a scale factor of $\frac{7}{3}$,we must divide the ray $BX$ into $7$ equal parts (since $7 > 3$).
Steps of construction:
$1.$ Draw a ray $BX$ making an acute angle with $BC$.
$2.$ Locate $7$ points $B_{1}, B_{2}, \dots, B_{7}$ on $BX$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$3.$ Join $B_{3}$ to $C$ (since the denominator is $3$).
$4.$ Draw a line through $B_{7}$ parallel to $B_{3}C$ intersecting the extended line segment $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting the extended line segment $BA$ at $A'$.
The given statement claims that $B_{6}C'$ is drawn parallel to $B_{3}C$,which is incorrect because the scale factor is $\frac{7}{3}$,requiring the parallel line to be drawn from $B_{7}$,not $B_{6}$.