AB is an arc of a circle with the centre O, $\angle AOB = \pi/2$ radians and $OA = 5$ cm. AC is parallel to OB and BC is an arc of a circle with centre A.

1. Find the radius of the circle centred at $A$ for arc $BC$

Since $AB$ is an arc of a circle with centre $O$ and $OA=5$ cm, $OB=5$ cm.

Triangle $AOB$ is right-angled at $O$, so
$AB=\sqrt{OA^2+OB^2}=\sqrt{5^2+5^2}=5\sqrt{2}\text{ cm}.$

Since arc $BC$ has centre $A$, its radius is $AB=5\sqrt{2}$ cm.

2. Find the angle subtended by arc $BC$ at centre $A$

As $AC\parallel OB$ and $AO\perp OB$, $AC\perp AO$.

In triangle $AOB$, $OA=OB$, so it is an isosceles right-angled triangle. Hence
$\angle OAB=\frac{\pi}{4}.$

Therefore
$\angle BAC=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.$

3. Use the arc length formula

Arc length $BC=r\theta$.

$BC=5\sqrt{2}\times \frac{\pi}{4}=\frac{5\sqrt{2}\pi}{4}=\frac{5\pi}{2\sqrt{2}}=\frac{5\pi}{\sqrt{2}}\text{ cm}.$