Find an example of a regular triangle-free $4$-chromatic graph
If you're lazy you can check all the graphs on Wikipedia. The Hoffman–Singleton graph, for example, is strongly regular, has girth 5, and has chromatic number 4.
Another example is given by Kneser graphs $K(n,k)$ with suitable parameters.
By Lovasz' theorem, the chromatic number of $K(n,k)$ is given by $n-2k+2$.
Moreover, if $n<3k$ we have that $K(n,k)$ is triangle-free, so:
$\color{red}{K(8,3)}$ is a triangle-free, $10$-regular graph on $56$ vertices with chromatic number $\chi=4$.
However, the minimal example of a triangle-free, regular graph with $\chi=4$ is given by another "topological graph", the Clebsch graph, with $16$ vertices and degree $5$.
Discussed this question with my guide after getting a specific answer by joining two Mycielskian of $C_5$ appropriately (so $22$ vertices). He told me the following technique, which can be generalized to handle such questions.
Construct the Mycielskian of $C_5$ (any triangle free $4$-chromatic graph will do). We have five vertices with degree $3$, five with degree $4$ and one with degree $5$. Make five copies of this. Add a new vertex and join to the corresponding degree $4$ vertex in each of the five copies. Do this for the rest degree $4$ vertices. For degree $3$ vertices take two vertices and join both of them to the corresponding degree $3$ vertex in each of the five copies. Now the final graph is regular, triangle free and $4$-chromatic.
This technique is already published (unable to find the paper).
ADDENDUM:
For this general method, you use the exact same coloring in all five copies ($a_i$,$b_i$,...,$e_i$ for $i = 1$ to $11$) of the Mycielskians (color 1 to 4). Every corresponding degree 4 vertex of each is connected to a new vertex (i.e. $a_1$,$b_1$,...,$e_1$, having same color say 1, is connected to a new vertex whose degree is $5$ which can be assigned a different color than 1). Do this for the rest degree 4 vertices ($a_i$,$b_i$,...,$e_i$ for $i = 2$ to $5$). Every corresponding degree 3 vertex of each is connected to two new vertices (i.e. $a_6$,$b_6$,...,$e_6$, having same color say 1, is connected to two new vertices whose degree is $5$ which can be assigned a different color than 1). Do this for the rest degree 4 vertices ($a_i$,$b_i$,...,$e_i$ for $i = 7$ to $10$).
This results in a regular $\Delta$-free $4$-chromatic graph.
PARTICULAR ANSWER: (not using the general method described above which would result in a graph of order $70$)
The following is a particular answer to this question using Mycielskian resulting in a graph of order $22$. These are two copies of Mycielskian of $C_5$. I have not drawn all edges within each Mycielskian to improve the clarity. The left 5 vertices of Mycielskian were degree 4 vertices hence have only one matching, where as the middle five vertices of Mycielskian were degree 3 vertices hence have two matchings to make them regular.
