(b)
Reflexive $xRx$
Since, $x^{2} = x.x$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} = xy$
Transitive, $xRy \Rightarrow x^{2} = xy$
And $yRz \Rightarrow y^{2} = yz$
Now, $x^{2}y^{2} = xy^{2}z \Rightarrow x^{2} = xz$
$\Rightarrow xRz$
$\therefore$ It is transitive.