= Solution
Choose $z$ normal to the sheet and $x$ along the in-plane projection of the rotation axis. A uniform square has $I_x=I_y=ma^2/12$ and $I_z=ma^2/6$. Since $\omega_x=\omega_z=\omega/\sqrt2$, the two nonzero angular-momentum components are $J_x=I_x\omega/\sqrt2$ and $J_z=I_z\omega/\sqrt2$. Therefore
$$
\boxed{|\mathbf J|=\frac{ma^2\omega}{12}\sqrt{\frac52}=\frac{\sqrt{10}}{24}ma^2\omega.}
$$
<Angular momentum> is not parallel to the prescribed rotation axis. Multiplying the <moment of inertia> \i[about that axis] by $\omega$ would give the projection of <angular momentum> on the axis, not its magnitude.
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