As a supplement to Ivo Beckers's answer, I provide an explicit construction of Professor Halfbrain's cube:

The cube has edge length $10\sqrt{21}$. The angles between the three bottom edges and the table plane are (colored red, green, and blue, respectively):
$$
\color{red}{\sin^{-1}\left(\frac{1}{\sqrt{21}}\right)} \\
\color{lime}{\sin^{-1}\left(\frac{2}{\sqrt{21}}\right)} \\
\color{blue}{\sin^{-1}\left(\frac{4}{\sqrt{21}}\right)}
$$
As Ivo conjectures, this arrangement is unique (up to reflection and rotation around the vertical axis). We can prove this by using Euler angles to describe the rotation of the cube. (I will use the z-y-z convention.) We can ignore the first rotation about the z-axis, since it does not affect the angles between the cube and the table surface, only spinning the cube about its vertex. Calling the remaining two angles $\alpha$ and $\beta$, we can write the transformation matrix:
$$
\left(\begin{matrix} \cos\alpha\cos\beta & -\cos\alpha\sin\beta &\sin\alpha\\
\sin\beta &\cos\beta & 0\\
-\sin\alpha\cos\beta &\sin\alpha\sin\beta &\cos\alpha\\\end{matrix}\right)
$$
and its application to the three unit vectors:
$$
\left(
\begin{matrix}
\cos\alpha\cos\beta\\
\sin\beta\\
-\sin\alpha\cos\beta\\
\end{matrix}
\right),\left(
\begin{matrix}
-\cos\alpha\sin\beta\\
\cos\beta\\
\sin\alpha\sin\beta\\
\end{matrix}
\right),\left(
\begin{matrix}
\sin\alpha\\
0\\
\cos\alpha\\
\end{matrix}
\right)
$$
With this we have the height of three vertices of the cube, as well as the heights of the three edges from those vertices to the bottommost vertex. The position of each vertex is a sum of some subset of those edges.
The height of the shortest edge must be $10$; otherwise—as the height of each edge is positive—no vertex could have height $10$. Next, since each of the three different heights may be included at most once, the next edge must have height $20$, otherwise no vertex could have height $20$. Similarly, the third edge must have a height of $40$. This gives us enough to reach all the way to $10+20+40=70$; and, since our choice of heights were fixed at each step, we know the solution must be unique.
Using $s$ as the edge length, we can now set up a system of equations to solve for the Euler angles and size of the cube:
$$
\begin{align}
-s\sin\alpha\cos\beta &= 10 \\
s\sin\alpha\sin\beta &= 20 \\
s\cos\alpha &= 40
\end{align}
$$
Immediately we can write $s=40\sec\alpha$ and remove $s$ from the first two equations:
$$
\begin{align}
-4 \cos \beta \tan \alpha &= 1 \\
2\sin \beta \tan \alpha &= 1
\end{align}
$$
Then divide and write:
$$
-\frac{1}{2}\tan\beta = 1 \\
\beta = -\tan^{-1} 2
$$
Then we can back-substitute and solve for $\alpha$:
$$
-\frac{4}{\sqrt{5}}\tan\alpha = 1 \\
\alpha = -\tan^{-1}\left(\frac{\sqrt{5}}{4}\right)=\cos^{-1}\left(\frac{4}{\sqrt{21}}\right)
$$
Finally, using $s=40\sec\alpha$ we then obtain:
$$
s = 10\sqrt{21}
$$
For completeness, here is the resulting rotation matrix:
$$
\left(
\begin{matrix}
\frac{4}{\sqrt{105}} & \frac{8}{\sqrt{105}} & -\sqrt{\frac{5}{21}} \\
-\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 0 \\
\frac{1}{\sqrt{21}} & \frac{2}{\sqrt{21}} & \frac{4}{\sqrt{21}} \\
\end{matrix}
\right)
$$