The impulse of the net force is equal to the change in momentum, $\vec{I} = \Delta\vec{p} = m(\vec{v}_f - \vec{v}_i)$. In uniform circular motion, the speed $v$ is constant. Over a quarter period, the velocity vector rotates by 90°, so the initial velocity $\vec{v}_i$ and final velocity $\vec{v}_f$ are perpendicular. The magnitude of the change in velocity is $|\Delta\vec{v}| = |\vec{v}_f - \vec{v}_i| = \sqrt{v_f^2 + (-v_i)^2} = \sqrt{v^2+v^2} = v\sqrt{2}$. The magnitude of the impulse is therefore $|I| = m|\Delta v| = mv\sqrt{2}$.