= Solution
For a general positive side length $s$, the cube has <volume> $s^3$, while separability of the <integral> gives
$$
\int_0^s\int_0^s\int_0^sxyz\,dx\,dy\,dz
=\left(\frac{s^2}{2}\right)^3=\frac{s^6}{8}.
$$
Thus the requested general dependence of the mean <temperature> is
$$
\boxed{\overline T(s)=\frac{s^3}{8}.}
$$
For the cube with side length three this gives
$$
\boxed{\overline T(3)=\frac{27}{8}=3.375.}
$$
Equivalently each coordinate has volume-average value $s/2$, and the rectangular integration domain makes the mean of $xyz$ the product $(s/2)^3$.
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