Even if a satisfying answer has been found by the asker, I want to show how the answer can be reached without matrices nor calculus.

If we take a plane surface with area $A$ and unit normal $\vec n$, then the projection of that area along a direction $\vec v$ (with $|\vec v|=1$) is $$ A'=A \vec n\cdot\vec v. $$ Let's take now a convex polyhedron: if the normals $\vec n_k$ of its faces are oriented outwards, then only those faces with $\vec n_k\cdot\vec v>0$ are visible from direction $\vec v$. The area projected by the polyhedron is then $$ S=\sum_{\text{$k$ visible}} A_k \vec n_k\cdot\vec v, $$ where $A_k$ is the area of the $k$-the face, and the sum is only carried over the visible faces.

In the case of a parallelepiped, we can set up our coordinates such that its center is at the origin, and its faces are parallel to coordinate planes, so that $\vec n$ can take one of the six values $(0,0,\pm1)$, $(0,\pm1,0)$,$(\pm1,0,0)$. Without loss of generality, we can in addition suppose that $v=(x,y,z)$ is in the first octant. We get then: $$ \tag{1} S=A_x x+A_y y+A_z z, $$ where $A_x$ is the area of the faces perpendicular to $x$ axis, and so on.

$S$ is the quantity to be maximised, subject to the constraint $$ \tag{2} x^2+y^2+z^2=1. $$ But $(1)$ is the equation of a plane (if $S$ is fixed) and $(2)$ is the equation of a sphere. We obtain the maximum value of $S$ when the plane is tangent to the sphere, i.e. when $v$ is perpendicular to the plane, leading to: $$ v={(A_x,A_y,A_z)\over\sqrt{A_x^2+A_y^2+A_z^2}}. $$ Substituting this into $(1)$ we can find the maximum value of $S$: $$ S_\max=\sqrt{A_x^2+A_y^2+A_z^2}. $$ For the case at hand, substituting: $$ A_x=A_y=0.03\text{ m}^2,\quad A_z=0.01\text{ m}^2, $$ we get $S_\max={\sqrt{19}\over100}\approx0.0436\text{ m}^2$.

Even if a satisfying answer has been found by the asker, I want to show how the answer can be reached without matrices nor calculus.

If we take a plane surface with area $A$ and unit normal $\vec n$, then the projection of that area along a direction $\vec v$ (with $|\vec v|=1$) is $$ A'=A \vec n\cdot\vec v. $$ Let's take now a convex polyhedron: if the normals $\vec n_k$ of its faces are oriented outwards, then only those faces with $\vec n_k\cdot\vec v>0$ are visible from direction $\vec v$. The area projected by the polyhedron is then $$ S=\sum_{\text{$k$ visible}} A_k \vec n_k\cdot\vec v, $$ where $A_k$ is the area of the $k$-the face, and the sum is only carried over the visible faces.

In the case of a parallelepiped, we can set up our coordinates such that its center is at the origin, and its faces are parallel to coordinate planes, so that $\vec n$ can take one of the six values $(0,0,\pm1)$, $(0,\pm1,0)$,$(\pm1,0,0)$. Without loss of generality, we can in addition suppose that $v=(x,y,z)$ is in the first octant. We get then: $$ \tag{1} S=A_x x+A_y y+A_z z, $$ where $A_x$ is the area of the faces perpendicular to $x$ axis, and so on.

We must find for which $v=(x,y,z)$ the value of $S$ is maximum, with the constraint $|\vec v|=1$, that is: $$ \tag{2} x^2+y^2+z^2=1. $$ But $(1)$ is the equation of a plane (if $S$ is fixed) and $(2)$ is the equation of a sphere. We obtain the maximum value of $S$ when the plane is tangent to the sphere, i.e. when $v$ is perpendicular to the plane, leading to: $$ v={(A_x,A_y,A_z)\over\sqrt{A_x^2+A_y^2+A_z^2}}. $$ Substituting this into $(1)$ we can find the maximum value of $S$: $$ S_\max=\sqrt{A_x^2+A_y^2+A_z^2}. $$ For the case at hand, substituting: $$ A_x=A_y=0.03\text{ m}^2,\quad A_z=0.01\text{ m}^2, $$ we get $S_\max={\sqrt{19}\over100}\approx0.0436\text{ m}^2$.