Given in the question,
$\angle BOC={{45}^{\circ }}$
$\angle AOC=180-45={{135}^{\circ }}$[ linear pair]
Area of circle $=\pi {{r}^{2}}$
Area of region $x=\theta /360\times \pi {{r}^{2}}$
$=135/360\times \pi {{r}^{2}}$
$=3/8\times \pi {{r}^{2}}$
The probability that the spinner will land in the region
X= area of region x/ total area of circle
$x=\frac{\frac{3}{8}\pi {{r}^{2}}}{\pi {{r}^{2}}}$
$x=3/8$
$x=3/8$
Therefore, the probability that the spinner will land in region X is $3/8$.