As an abelian group, there is an isomorphism $\mathbb R\cong\mathbb Q^{(I)}$ with $I$ a set of cardinality equal to that of $\mathbb R$; here $\mathbb Q^{(I)}$ denotes a direct sum of $|I|$ copies of $\mathbb Q$, as usual. This is a consequence of the fact that $\mathbb R$ is a $\mathbb Q$-vector space, and it has a basis as such.
Since the tensor product distributes over arbitrary direct sums, $\mathbb R\otimes_{\mathbb Z}\mathbb R\cong \mathbb Q^{(I)}\otimes_{\mathbb Z}\mathbb Q^{(I)}\cong(\mathbb Q\otimes \mathbb Q)^{(I\times I)}.$ Now, since $I\times I$ has the cardinality of $I$ and you know that $\mathbb Q\otimes \mathbb Q\cong\mathbb Q$, we have $(\mathbb Q\otimes \mathbb Q)^{(I\times I)}\cong\mathbb Q^{(I)}\cong\mathbb R$.