In computing $$ \newcommand{\Z}{\mathbb{Z}} \newcommand{\op}{\oplus} \Bigl( \Z^2 \op (\Z/6\Z) \op (\Z/126\Z) \Bigr) \otimes_\Z \Bigl( \Z \op (\Z/45\Z) \op (\Z/495\Z) \Bigr), $$ do I get as exponent of $\Bbb Z$ in the product, $3$ or $2$? The ring $\Bbb Z$ should be the identity element so I think $2$, but distributivity says:
is it $\Bbb Z ^2 \otimes \Bbb Z= (\Bbb Z\oplus \Bbb Z)\otimes \Bbb Z$
or
is it $\Bbb Z ^2 \otimes \Bbb Z = (\Bbb Z \otimes \Bbb Z)\otimes \Bbb Z$?
Exponents are pretty much never used like this to denote "powers" with respect to the tensor product. If you wanted to write $\mathbb{Z}\otimes\mathbb{Z}$ as a "power", the standard way to do that would be $\mathbb{Z}^{\otimes 2}$ with the little $\otimes$ in the exponent to indicate you mean a tensor power.
Instead, if there is no context that indicates otherwise, an exponent on any sort of algebraic structure normally indicates a power with respect to the Cartesian product, so $\mathbb{Z}^2$ means $\mathbb{Z}\times\mathbb{Z}$. In the case of abelian groups if the exponent is finite, this is equivalent to a direct sum as well.