All the information on the higher ramification groups can be derived from Theorem II.5.6, which TKe notes.

The Galois group of the ray class field is the group of fractional ideals relatively prime to $\mathfrak p$ modulo the principal ideals with generators $1$ mod $\mathfrak p$. This naturally maps to the class group, and the kernel consists of principal ideals, and is hence isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$. The ramification group at $\mathfrak p$ in the abelianization of the Galois group is $\mathcal O_{K_p}^\times$, and the maps is $\mathcal O_{K_p}^\times \to (\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times$ is the obvious projection factoring through $(\mathcal O_K/\mathfrak p)^\times$.

Hence the ramification group of the extension is isomorphic to $(\mathcal O_K/\mathfrak p)^\times / \mathcal O_K^\times $ and is entirely tame - there are no higher ramification groups.

$\mathfrak p$ is ramified if and only if the natural map $ \mathcal O_K^\times \to (\mathcal O_K/\mathfrak p)^\times$ is not surjective.

This map can only be surjective when $\left|(\mathcal O_K/\mathfrak p)^\times \right| \leq \left| \mathcal O_K^\times\right|$, which happens:

if $K\neq \mathbb Q(i), \mathbb Q(\mu_3)$, only when $\mathfrak p$ is a split or ramified prime lying over $2$ or $3$.

If $K = \mathbb Q(i)$, only when $\mathfrak p$ is the prime lying over $2$ or one of the two primes lying over $5$.

If $K = \mathbb Q(\mu_3)$, only when $\mathfrak p = 2$, the prime lying over $3$, or one of the two primes lying over $7$.

In all these cases, it is easy to see that the map is indeed surjective and so $\mathfrak p$ does not ramify.

Advanced Topics in the Arithmetic of Elliptic Curves, Theorem II.5.6 + Theorem II.3.2(c): The conductor of $K(j(E),h(E[\mathfrak{c}]))/K$divides$\mathfrak{c}$. This gives a supset of the ramified primes. $\endgroup$ – user19475 Jan 26 '18 at 6:01