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Second try: Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let ${k_v}'$ be the subfield of $k_v$ consisting of elements that are algebraic over $k$ and lie in $\mathbb{Q}^{ab}$ (under any choice of embedding), and let $m_v$ be a positive integer such that ${k_v}' \subset \mathbb{Q}[\zeta_{m_v}]$. Let $m_w$ be defined similarly. Let $n$ be the least common multiple of $m$, $m_v$, and $m_w$, and let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over the subfield of $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

Okay, I'm giving up. I hadn't realized local fields had large subfields of cyclotomic extensions. Thanks to BCnrd - I learned something interesting today.

Second try (also wrong): Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let ${k_v}'$ be the subfield of $k_v$ consisting of elements that are algebraic over $k$ and lie in $\mathbb{Q}^{ab}$ (under any choice of embedding), and let $m_v$ be a positive integer such that ${k_v}' \subset \mathbb{Q}[\zeta_{m_v}]$. Let $m_w$ be defined similarly. Let $n$ be the least common multiple of $m$, $m_v$, and $m_w$, and let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over the subfield of $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

Second try: Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let $n$ be the least common multiple of $\mu(k_v)$, $\mu(k_w)$, and $m$. Let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

Second try: Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let ${k_v}'$ be the subfield of $k_v$ consisting of elements that are algebraic over $k$ and lie in $\mathbb{Q}^{ab}$ (under any choice of embedding), and let $m_v$ be a positive integer such that ${k_v}' \subset \mathbb{Q}[\zeta_{m_v}]$. Let $m_w$ be defined similarly. Let $n$ be the least common multiple of $m$, $m_v$, and $m_w$, and let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over the subfield of $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?

Second try: Choose an embedding of $k$ in an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $k' = k \cap \mathbb{Q}^{ab}$, where $\mathbb{Q}^{ab} \subset \overline{\mathbb{Q}}$ is the maximal abelian extension. By Kronecker-Weber, there exists a positive integer $m$ such that $k' \subset \mathbb{Q}[\zeta_m]$. Let $n$ be the least common multiple of $\mu(k_v)$, $\mu(k_w)$, and $m$. Let $K = k[\zeta_n]$.

$K$ is an abelian Galois extension of $k$ and $\mu(K) = n$. The minimal polynomial of $\zeta_n$ over $k$ splits into (Galois conjugate) polynomials of equal degree over $k_v \cap \overline{\mathbb{Q}}$, so $K \otimes_k k_v$ is isomorphic to a product of copies of $k_v[\zeta_n]$. Therefore, for any place $v'$ over $v$, $K_{v'} \cong k_v[\zeta_n]$ and $\mu(K_{v'}) = n$. Similarly, $\mu(K_{w'}) = n$ for any place $w'$ over $w$.

First Try (wrong - see BCnrd's comment): Let $n$ be the least common multiple of $\mu(k_v)$ and $\mu(k_w)$, let $K = k[\zeta_n]$, where $\zeta_n$ is a primitive $n$th root of unity, and let $v'$ and $w'$ be any chosen places over $v$ and $w$, respectively. $K$ is Galois over $k$, and if I'm not mistaken, we have $\mu(K_{v'}) = \mu(k_v[\zeta_n]) = n$ and $\mu(K_{w'}) = \mu(k_w[\zeta_n]) = n$. Is there a subtlety I'm missing?