Let the original mass of the ring be $M$ and its radius be $R$. The moment of inertia of the complete ring about an axis passing through its centre and perpendicular to its plane is $I_{total} = MR^2$. An arc corresponding to a $90^\circ$ sector represents one-fourth $\left(\frac{90^\circ}{360^\circ} = \frac{1}{4}\right)$ of the complete ring. Therefore, the moment of inertia of the removed portion about the same axis is $I_{removed} = \frac{1}{4}MR^2$. The moment of inertia of the remaining part is $I_{remaining} = I_{total} - I_{removed} = MR^2 - \frac{1}{4}MR^2 = \frac{3}{4}MR^2$. Given that the moment of inertia of the remaining part is $K$ times $MR^2$, we have $K = \frac{3}{4}$.