Writing $\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_I \mathbb{Q}$ where $I$ indexes a Hamel basis for $\mathbb{R}$ minus one element, we have

$$\text{Ext}^1(\mathbb{R}/\mathbb{Z}, \mathbb{Z}) \cong \text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \times \prod_I \text{Ext}^1(\mathbb{Q}, \mathbb{Z}).$$

At this point we can write $\mathbb{Q}/\mathbb{Z}$ as the direct product of its Sylow subgroups, but we can also observe the following: there is a short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$ giving rise to a long exact sequence many of whose terms vanish, namely

$$0 \to \text{Hom}(\mathbb{Z}, \mathbb{Z}) \to \text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \to \text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \to 0 \to \dots$$

establishing that $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ is the quotient of $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ by a copy of $\mathbb{Z}$, so to compute the former it suffices to compute the latter (and figure out what copy of $\mathbb{Z}$ comes into play).

Now we can compute $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ by writing $\mathbb{Q}/\mathbb{Z}$ as a filtered colimit of its subgroups $\frac{1}{n} \mathbb{Z}/\mathbb{Z}$ and using that $\text{Ext}^1$ sends filtered colimits in the first argument to cofiltered limits. (**Edit, 10/10/20:** Excuse me, this is false. There is a $\lim^1$ term involving $\lim^1 \text{Hom}(\frac 1 n \mathbb{Z}/\mathbb{Z}, \mathbb{Z})$ which vanishes.) This gives that

$$\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, \mathbb{Z}) \cong \widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$

is the profinite integers. Now it's very tempting to conjecture that the copy of $\mathbb{Z}$ we need is the obvious one, giving

$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \widehat{\mathbb{Z}}/\mathbb{Z}.$$

But this description is somewhat unsatisfying, for the following reason: by functoriality $\text{Ext}^1(\mathbb{Q}, -)$ always takes values in $\mathbb{Q}$-modules, which is to say $\mathbb{Q}$-vector spaces, but we've written this $\mathbb{Q}$-vector space as a quotient of two things which are not $\mathbb{Q}$-vector spaces. We can rectify this a bit by tensoring the above by $\mathbb{Q}$, which fixes it, giving

$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \left( \widehat{\mathbb{Z}} \otimes \mathbb{Q} \right) / \mathbb{Q}.$$

$\widehat{\mathbb{Z}} \otimes \mathbb{Q}$ has another name: it is the ring of finite rational adeles $\mathbb{A}_{\mathbb{Q}}$. This is the form in which the answer is stated in these notes. This is a $\mathbb{Q}$-vector space of dimension the reals and so another amusing way to state the answer is that

$$\text{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R}.$$

At this point we don't really need to know what copy of $\mathbb{Z}$ we need to quotient $\widehat{\mathbb{Z}}$ by.

Altogether we get, abstractly, that $\text{Ext}^1(\mathbb{R}/\mathbb{Z}, \mathbb{Z})$ is the product of $\widehat{\mathbb{Z}}$ and a $\mathbb{Q}$-vector space $\prod_I \mathbb{R}$ of some very large dimension ($2^{2^{\aleph_0}}$?).

Incidentally, although it's not needed for this computation, the same filtered colimit argument as above gives that

$$\text{Ext}^1(\mathbb{Z}(p^{\infty}), \mathbb{Z}) \cong \mathbb{Z}_p.$$