In every square in the $3 \times 4$ rectangle put the number $\frac{45}{4}$. In the $3\times 5$ rectangle put $3$. In the $4\times 5$ rectangle put $\frac{21}{4}$.

That is not working when circling the box by $6\cdot 3 + 8\cdot \dfrac {21} {4} = 60 \neq 120$. However, replacing $\dfrac {45} {4}, 3, \dfrac {21} {4}$ with the values $5, 8,9$ will work.

More generally, if we set the values on each pair of opposite faces to be all the same, we're just solving a system of three linear equations in three unknowns that happens to be obviously nondegenerate, i.e., there always exists a unique solution. Is this really the intended question?

Yeah... I must have set up the equations wrongly by accidentally dividing 120 by 2 twice.