reduce k-colorable to 3-colorable graph problem
It is known that K-Colorability reduces to 3-Colorability in polynomial time as obviously both problems are NP-complete, and NP-complete problems are polynomial-time reducible to each other. However, in order to $k$-color a graph using a 3-coloring oracle, you need a specific reduction from $k$-colorability to $3$-colorability which is constructive in the sense that given a 3 coloring of the output graph, then you have a way to extract from it a $k$-coloring of the input graph (and not only prove that there exists a $k$-coloring). Such a reduction, was shown by László Lovász  here. Specifically, it was shown that Colorability reduces to 3-Colorability in polynomial time which is sufficient to solve your problem using the same reduction. Here is a very good explanation by Vašek Chvátal's for Lovász's reduction. The reduction has the following format:  given an instance to the Colorability problem which is a pair $\langle k, G \rangle $, where $k$ is a positive integer and $G$ is an undirected graph, the reduction outputs in polynomial time, a graph $G'$ such that $G'$ has a 3-coloring iff $G$ has a k-coloring. The proof in the link is actually constructive, in particular, it details how a k-coloring of $G$  can be easily extracted from a 3-coloring of $G'$ (When I say easily, I mean in polynomial time). So to solve your problem: given a graph $G$, you can consider the pair $x = \langle k, G\rangle$ ($k$ is constant in your problem) and feed $x$ as an input to the reduction to get $G'$ in polynomial time. Now using your oracle, you can 3-color $G'$, and then use the 3-coloring of $G'$ to extract a $k$-coloring of $G$.