Based on your sketch, the block is always moving along it's long axis, in other words, the velocity is always along the direction of your red vector. This means in the picture on the left, there is only vertical velocity while in the picture on the right, there is both vertical and horizontal velocity. This is what you have described, I am only summarizing to make sure the rest of my answer makes sense and is consistent with this understanding.
Now, you claim the drag coefficient of the case on the left is known yet the drag coefficient for the case on the right is unknown. To clarify a point, the drag coefficient is a scalar, not a vector as you claimed it to be. And in this case, the drag coefficient is actually the same in both cases if the block is moving along the vectors shown. Why is this the case?
Consider the area vector of each face to be the area of the face times the surface normal, ie.:
$$\vec{A} = A\cdot\hat{n}$$
In your example on the left, the area normal is in the $\hat{j}$ or Y-normal direction and so is the velocity. This gives you the drag coefficient you cite.
In your example on the right, you now have a component of the area vector in the $\hat{i}$ direction and a component in the $\hat{j}$ direction. But your velocity also has components in both those directions.
You are correct to say that the area increases in the $\hat{j}$ direction -- you now have both a portion of the short side and a portion of the long side exposed to the Y-direction. However, the block is still moving normal to the short side along the red vector, which means the area vector for the long sides are perpendicular to the red vector and that area does not contribute to the so called "frontal area" of the block.
Yes you could. Since the force on an object from drag is given by
$$F_D = \frac{1}{2}\rho v^2 A C_D$$
where $C_D$ is the drag coefficient, then all you would need to know are your velocity ($v$), your fluid density $\rho$, your cross sectional area ($A$) and the force of gravity on the body, which would then be equivalent to $F_D$ since the body would have no net force on it. Thus, you could isolate for $C_D$, getting
$$C_D = 2mg/\rho v^2 A$$
Best Answer
According to Sighard Hoerner's Fluid Dynamic Drag, this would be the half-sphere with the open side exposed to the wind. Its drag coefficient is 1.42. A rod with a hemispherical cross section will even have a drag coefficient of 2.3 (right column in the graph below).
If you restrict the competition to solid objects, still the half sphere wins with a drag coefficient of 1.17. In all cases, the reference area is the cross section orthogonal to the flow direction.
Figure 33 from Sighard Hoerner's Fluid Dynamic Drag, Chapter 3.
Note that the difference in drag of half spheres due to their orientation is used in anemometers for measuring the wind speed. When the open face is turned away from the wind, its drag coefficient drops to 0.42.
The reason for the difference, and the high drag when the open side is exposed to the wind, is the massive separation around and behind the sphere. Air flowing out from the inside and over the rim of the sphere will need some space to "turn around", effectively increasing the blocked cross section that the outside flow experiences. When the round side is exposed to the wind, the separation is restricted to the cross section of the sphere itself.