Let $k$ be a number field, $S$ a set of places of $k$, $k_S$ the maximal extension of $k$ unramified outside $S$, $\mathcal{O}_S$ the subring of $k_S$ with $\nu_{\mathfrak{p}}(\alpha)\geq 0$ for all $\mathfrak{p}\not\in S$, and $G_{S}$ the galois group $\text{Gal}(k_S/k)$. I am considering the following exercise at the end of chapter 8.3 in Cohomology of Number Fields.
Show the isomorphism $H^2(G_S,\mathcal{O}_S^\times)\cong Br(k_S/k)$
My attempt was to consider the following exact sequence
$$0\longrightarrow k_S^\times \longrightarrow I_{k_S}\longrightarrow C_{k_S}\longrightarrow 0$$
Where $I_{k_S}$ and $C_{k_S}$ are the Ideles and Idele class group of $k_S$. (Note that if $k_S$ is an infinite extension, these are simply the Galois invariance of the absolute Ideles and Idele class group.)
We take the cohomology long exact sequence of this exact sequence and obtain at $H^2$:
$$0\simeq H^1(G_S, C_{k_S})\longrightarrow H^2(G_S,k_S^\times)\simeq Br(k_S/k)\longrightarrow H^2(G_S,I_{k_S})\longrightarrow H^2(G_S, C_{k_S})$$
This gives an isomorphism $Br(k_S/k)\simeq \ker(H^2(G_S,I_{k_S})\to H^2(G_S, C_{k_S}))$. Now suppose $k=\mathbb{Q}$ and $S=\{3,\infty\}$, so $3^\infty|\#G_S$. Since $C$ is a formation module (defn. 3.1.8 in Cohomology of Number Fields) we have $H^2(G_S,C_{k_S})\simeq \frac{1}{\#G_S}\mathbb{Z}/\mathbb{Z}$. On the other hand, for $H^2(G_S,I_{k_S})$ we have
$$H^2(G_S,I_{k_S})(3)\simeq H^2(k,I)(3)\simeq \bigoplus_{\mathfrak p} H^2(k_{\mathfrak p})(3)\simeq \bigoplus_{\mathfrak p} \mathbb{Q}_3/\mathbb{Z}_3$$
by propositions $8.1.8$ and $8.1.7$ of the same book. (Note that we are using $A(p)$ to denote the $p$-power torsion of $A$.) Altogether, this yields
$$Br(k_S/k)(3)=H^2(G_S,k_S^\times)(3)\simeq \ker \left(\bigoplus_{\mathfrak p} \mathbb{Q}_3/\mathbb{Z}_3\longrightarrow \mathbb{Q}_3/\mathbb{Z}_3\right)$$
where the map is given by summation. Now by 8.3.11 of the same book, we have
$$H^2(G_S,\mathcal{O}_S^\times)(p)\simeq \ker\left(\bigoplus_{\mathfrak p\in S\setminus S_\infty} \mathbb{Q}_3/\mathbb{Z}_3\longrightarrow \mathbb{Q}_3/\mathbb{Z}_3\right)=\ker(\mathbb{Q}_3/\mathbb{Z}_3\to \mathbb{Q}_3/\mathbb{Z}_3)\simeq 0.$$
So these are two formulas, one for the $3$-part of $Br(k_S/k)$ and another for the $3$-part of $H^2(G_S,\mathcal{O}_S^\times)$. One is trivial and the other is decidedly not trivial, so they cannot be isomorphic.
Here is another approach I took, which similarly suggested that the groups are not isomorphic, or at least not canonically isomorphic. We have a canonical embedding $\mathcal{O}_S^\times\to k_S^\times$, and we would hope the isomorphism of their Brauer groups is induced by this embedding. To that end, suppose for simplicity that $k=\mathbb{Q}$ and $S=\{2,5,7,\dots,\infty\}$. we consider the following exact sequence
$$0\longrightarrow \mathcal{O}_S^\times \longrightarrow k_S^\times \longrightarrow \prod_{\mathfrak{p}|3} \mathfrak{p}^{\mathbb{Z}}=A$$
where the product is over the primes $\mathfrak{p}$ lying over $3$ and the last map need not be surjective. To turn this into a short exact sequence, we would like to determine the image of the last map.
To do this, fix a prime $\mathfrak{p}^*$ lying over $3$, and let $D\subseteq G_S$ be its decomposition group. The $A$ can be described as the set of functions $f:G_S\to \mathbb{Z}$ such that $f(\sigma\tau)=f(\tau)$ for all $\sigma \in D$. We identify each prime $\mathfrak{p}$ in the product with a coset of $G/H$, and then for $a\in A$ we set $f_a(H\sigma)=\nu_{(\mathfrak{p}^*)^\sigma}(a)$.
To determine the image of $k_S$ in $A$, note that anything in the image must be a continuous function. To see this, observe that any $\alpha \in k_S$ is in a Galois subextension $k_S/K/k$, and $f_\alpha$ then factors through the finite quotient $\text{Gal}(K/k)$ and is therefore continuous by properties of the Krull topology. Furthermore, all such continuous functions are possible: if $f\in A$ is continuous, then it factors through some finite quotient of $G_S$, and we can pass to the corresponding extension of $k$ to find a suitable $\alpha$.
With this description of the image of $k_S^\times$ in $A$ in mind, it becomes clear that we have constructed an isomorphism
$$k_S^\times/\mathcal{O}_S^\times\simeq \text{Ind}^D_{G_S}(\mathbb{Z})$$
since the collection of such continuous functions is precisely the definition of the induced module. Taking the long exact sequence in cohomology, we obtain
$$H^1(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\longrightarrow H^2(G_S,\mathcal{O}_S^\times)\longrightarrow H^2(G_S,k_S^\times)\longrightarrow H^2(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\longrightarrow H^3(G_S,\mathcal{O}_{S}^\times) $$
Computing the first and fourth term in this sequence, we obtain
$$H^1(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\simeq H^1(D,\mathbb{Z})\simeq H^1(\hat{\mathbb{Z}},\mathbb{Z})\simeq \varinjlim H^1(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})\simeq 0$$
$$H^2(G_S,\text{Ind}^D_{G_S}(\mathbb{Z}))\simeq H^2(D,\mathbb{Z})\simeq H^2(\hat{ \mathbb{Z}},\mathbb{Z})\simeq H^1(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\simeq \mathbb{Q}/\mathbb{Z}.$$
Taking the $p$-part of this exact sequence for $p\neq 3$ and using 8.3.11, we obtain
$$0\longrightarrow H^2(G_S,\mathcal{O}_S^\times)(p)\longrightarrow H^2(G_S,k_S^\times)(p)\longrightarrow \mathbb{Q}_p/\mathbb{Z}_p\longrightarrow 0.$$
Once again, we find that $H^2(G_S,\mathcal{O}_S^\times)$ is only a subobject of $H^2(G_S,k_S^\times)$, and similarly the failure for it to be an isomorphism comes from the primes not in $S$.
The ideas here apply more generally for other $S$ and $k$ and you still get a nontrivial cokernel of the inclusion in those cases.
I do not immediately see a flaw with either of these arguments, and the fact that they both show the exercise being false makes me think it might just be a mistake in the book, but I am far too aware of my shortcomings to believe this, so I am hoping someone can point out a mistake in these arguments.
Best Answer
I sent an email to Alexander Schmidt about this problem. It is indeed incorrect and the body of my post contains two counterexamples. He said he thinks the exercise should have been to prove that $H^2(G_S,k_S^\times)=br(k_S|k)$, but notes that this is a special case of theorem 6.3.4 of the same book.
The errata list has now been updated here.