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 $3/8$.