Note that the Galois group you are after is a subgroup of Gal$(\mathbb{F}_{q^k}/\mathbb{F}_{p})$ (if $q = p^m$). This is true for any tower $L/M/K$ where Galois groups are defined; Gal$(L/M)$ is always a subgroup of Gal$(L/K)$.
Now the Galois group Gal$(\mathbb{F}_{q^k}/\mathbb{F}_p)$ is generated by the Frobenius automorphism $Frob_p : x\mapsto x^p$. Which powers of this are in Gal$(\mathbb{F}_{q^k}/\mathbb{F}_q)$, i.e. which powers fix the field $\mathbb{F}_q$?
Well these are the powers of $Frob_p^m$. (check this)
So the Galois group you want is cyclic, generated by $Frob_p^m$. The claim about the degree follows easily.
EDIT: You do not need to worry about the extension being Galois (although if you know this then the claim about the size of the Galois group is trivial, by definition of Galois. This is the approach Benjamin Lim is using and is equally valid just slightly more theoretical).
Explicitly, we know that Gal$(\mathbb{F}_{q^k}/\mathbb{F}_p)$ is cyclic of order $mk$, generated by $Frob_p$. Thus the subgroup generated by $Frob_p^m$ has order $k$.
This agrees with:
$[\mathbb{F}_{q^k} : \mathbb{F}_q] = \frac{[\mathbb{F}_{q^k} : \mathbb{F}_p]}{[\mathbb{F}_{q} : \mathbb{F}_p]} = \frac{mk}{m} = k$.