A thin straight bar magnet of magnetic moment M is bent into an arc...

Let the initial length of the magnet be $L$ and its pole strength be $m$. The initial magnetic moment is $M = m \times L$.

When bent into an arc of a circle of radius $R$ subtending an angle $\theta$ (in radians) at the centre, the arc length is $L = R\theta \Rightarrow R = \dfrac{L}{\theta}$.

The new effective magnetic length $L^{\prime}$ is the straight-line distance (chord) between the poles:

$L^{\prime} = 2R \sin\left(\dfrac{\theta}{2}\right) = \dfrac{2L}{\theta} \sin\left(\dfrac{\theta}{2}\right)$

The new magnetic moment is $M^{\prime} = m \times L^{\prime} = m \times \dfrac{2L}{\theta} \sin\left(\dfrac{\theta}{2}\right) = M \left( \dfrac{2}{\theta} \sin\left(\dfrac{\theta}{2}\right) \right)$.

Given that $M^{\prime} = \dfrac{2\sqrt{2}M}{\pi}$, we have:

$\dfrac{2}{\theta} \sin\left(\dfrac{\theta}{2}\right) = \dfrac{2\sqrt{2}}{\pi}$

$\dfrac{\sin(\theta/2)}{\theta} = \dfrac{\sqrt{2}}{\pi}$

By checking standard angles, if $\theta = \dfrac{\pi}{2}$ ($90^{\circ}$):

$\dfrac{\sin(\pi/4)}{\pi/2} = \dfrac{1/\sqrt{2}}{\pi/2} = \dfrac{2}{\sqrt{2}\pi} = \dfrac{\sqrt{2}}{\pi}$.

Thus, the angle subtended at the centre is $90^{\circ}$.

Answer: $90^{\circ}$