The finite extensions of $\Bbb Q_p$ are also local fields, so you can use the usual characterization of local field extensions being unramified.
Recall: a finite extension of local fields $L/K$ is unramified iff $[L:K]=[\lambda:\kappa]$ where $\lambda,\kappa$ are the residue fields for $L,K$ respectively. This comes from the fundamental relationship that
$$e(L|K)f(L|K)=[L:K]$$
and the fact that $f(L:K)=[\lambda:\kappa]$. But then since your $K$ is a finite extension of $\Bbb Q_p$ you know the residue field is a finite extension of $\Bbb F_p$, so that unramified extensions are classified by intermediate fields $\ell$ such that
$$\kappa\subseteq \ell\subseteq \overline{\kappa}=\overline{\Bbb F_p}$$
the last equality since the algebraic closure of $\kappa$, being itself a finite extension of $\Bbb F_p$ is the same as the algebraic closure of $\Bbb F_p$. This completely classifies unramified extensions of such $K$. You even have a nice description: if $\ell\cong \Bbb F_{p^n}$ then the corresponding extension of $K$ is just $K(\mu_{p^n-1})$ with $\mu_k$ as usual the set of $k^{th}$ roots of $1$.
From this we may further deduce a nice corollary that the maximal, unramified extension of $K$ is simply $K(\mathbf{\mu})$ with
$$\mathbf{\mu}=\bigcup_{(n,p)=1,\; \mu_n\not\subseteq K} \mu_n$$
This correspondence is--as usual--due to Hensel's lemma and the characterization of finite fields as collections of roots of unity (plus $0$). Naturally the $(n,p)=1$ condition comes from the fact that all $p^{n}$ roots of unity are already in $\kappa$ since $1$ is the only one such and the Frobenius is an automorphism.
From this you can also see that since there is no $p^n$ roots of $1$ un an unramified extension, and since $\mu_{mn}=\mu_m\times\mu_n$ when $(n,m)=1$, that any extension of $K$ by $p^n$ roots of $1$ must be totally ramified, unless $K$ already contains those roots of $1$. Wedge this against the fact that you can always write any extension as an unramified, followed by a totally ramified, and you have your more general approach.
There aren't too many unramified extensions of a number field. In fact, if $K$ is a number field and $K^{unr}$ is the maximal unramified extension of $K$, then $G(K^{unr}/K)\simeq Cl(K)$, where $Cl(K)$ is the ideal class group of $K$. So, for instance, there are no unramified extensions of $\mathbb{Q}$. The inverse of this map behaves like $(\mathbb{Z}/n\mathbb{Z})^{\times}\rightarrow G(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, in that we send the equivalence class of a prime $[\mathfrak{p}]$ into the Frobenius of $\mathfrak{p}$. This means that a prime will split completely in $K^{unr}$ if and only if it is principle. It is possible to compute these fields, and sometimes it's easy.
If we're okay with some ramification, say ramification at all primes dividing the ideal $\mathfrak{m}$, then we can create the Ray Class Group of $\mathfrak{m}$, denoted $Cl_{\mathfrak{m}}(K)$ which is like the class group but it avoids the primes that divide $\mathfrak{m}$. We can then use a similar idea to create all extensions that are only ramified at primes dividing $\mathfrak{m}$. Ie, there is a field $K^{unr}_{\mathfrak{m}}$ that is the maximal field unramified at every prime not dividing $\mathfrak{m}$, we call this the Ray Class Field of $\mathfrak{m}$. If a prime power divides $\mathfrak{m}$, then we can control ramification at that prime.
There's a little more we can do. Sometimes a field can split over the real/complex numbers. We can control that by adding an additional, artificial term to $\mathfrak{m}$, making it a Modulus. We can then get Ray Class Fields of this modulus as well.
Class Field Theory then says that every single (abelian) field extension is a Ray Class Field of some modulus. That is good because we can compute ray class fields and categorize them. This paper gives a few algorithms that can compute these extensions.
Best Answer
I think that the answer-key to this question is the fundamental formula $n=ef$, where $n$ is the degree $[K':K]$ and $e$, $f$, are the ramification index and the residue-field extension degree, respectively.
To get all cube roots of unity into an extension of the residue field $\Bbb F_5$, you have to go to a field $\Bbb F_{5^m}$, whose multiplicative group is of order $5^m-1$, and cyclic, as I’m sure you know. Once $3|(5^m-1)$, you’ve got those cube roots of unity in your field. And of course $m=2$ serves very nicely, since $3|24$.
So by adjoining the cube roots of unity to $\Bbb Q_5$, you need a quadratic extension of the residue field. Thus here, $f=2$, forcing $e=1$. End of story.