= Solution
Put $J_0=4\sigma D^4/3=MD^2/3$. The normal to the lamina is already an <eigenvector>, with moment $3J_0$. The planar block has <characteristic polynomial>
$$
\det\begin{pmatrix}1-\lambda&-1\\-1&2-\lambda\end{pmatrix}=\lambda^2-3\lambda+1,
$$
so the <principal moments of inertia> are
$$
\boxed{I_-=\frac{J_0}{2}(3-\sqrt5),\qquad I_+=\frac{J_0}{2}(3+\sqrt5),\qquad I_3=3J_0.}
$$
A corresponding orthonormal set of <principal axes> is
$$
\boxed{\mathbf n_-=\frac{(2,\sqrt5-1,0)}{\sqrt{10-2\sqrt5}},\qquad\mathbf n_+=\frac{(2,-1-\sqrt5,0)}{\sqrt{10+2\sqrt5}},\qquad\mathbf n_3=(0,0,1).}
$$
The sign of each axis vector is immaterial. The smaller-moment axis makes angle $\tfrac12\arctan(2)$ with the positive $x_1$ direction, approximately $31.7^\circ$; the other planar axis is perpendicular. As a check, $I_-+I_+=3J_0=I_3$, the <perpendicular axis theorem> for a planar lamina.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2-lamina-principal-axes.png]
{title=The two planar <principal axes> of the sheared square lamina, with the third axis normal to the page}
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