Solution:

Consider a graph $G$ such that $\frac{a}{b} > \frac{1}{4}$. Note that the number of edges is $\frac{4a + 3b}{2}$. Additionally, if any two vertices of degree $4$ are adjacent, we can simply color that edge red and every other edge in $G$ white to get a valid quasi-coloring. Thus, suppose no two vertices of degree $4$ are adjacent. Then, consider the subgraph $G'$ resulting from removing all vertices of degree $4$. Note that $G'$ has $\frac{4a + 3b}{2} - 4a = \frac{3b - 4a}{2} < \frac{3b - b}{2} = b$ edges. Thus, $G'$ has fewer edges than vertices, which means that one of its connected components must be a tree $T$. In this tree $T$, we can just select some arbitrary vertex $v$, and perform a breadth-first search on $T$, greedily coloring the edges as we go along. What this means is that we partition the vertices of $T$ into disjoint subsets $S_0 = \{v\}, S_1, S_2, \ldots$, where $S_j$ contains all vertices of $T$ whose distance to $v$ is exactly $j$, and then we color edges between $S_0$ and $S_1$, and then greedily color the edges between $S_1$ and $S_2$, and so on. Once we color all the edges in $T$, we can color the edges between vertices in $T$ and the vertices of degree $4$ appropriately, and color the remaining edges white to get a valid quasi-coloring.

To show that $c = \frac{1}{4}$ is indeed the smallest possible solution, consider the graph shown below. This graph is not quasi-colorable (one can see that coloring the tails of this graph is impossible).

If there are $N$ vertices of degree $4$, there are $4N + 10$ vertices of degree $3$. Since $\frac{N}{4N + 10} \rightarrow \frac{1}{4}$ as $N \rightarrow \infty$, we have that for each value $c < \frac{1}{4}$, one can find a graph with $a = N_0$ vertices of degree $4$ and $b = 4N_0 + 10$ vertices of degree $3$ such that $\frac{a}{b} = \frac{N_0}{4N_0 + 10} > c$.

Solution:

Consider a graph $G$ such that $\frac{a}{b} > \frac{1}{4} \Longleftrightarrow b < 4a$. Note that the number of edges is $\frac{4a + 3b}{2}$. For each edge $e$ in the graph, create a variable $x_e$ in $\mathbb{F}_7$. Then, for each vertex $v$ of degree $3$, if the edges incident with $v$ are $e_1, e_2, e_3$, then consider the polynomial
$$P_v = x_{e_1}^2 + x_{e_2}^2 + x_{e_3}^2$$
in $\mathbb{F}_7$. There are $b$ such polynomials, each with total degree $2$. Thus, since $b < 4a$, we know that the sum of the total degrees over all polynomials, $2b$, is less than the total number of variables, $\frac{4a + 3b}{2}$. By Chevalley's Theorem, it follows that there exists a nontrivial solution to the system $P_v \equiv 0 \pmod{7}$ for all vertices $v$ of degree $3$. Consider one such solution. For each edge $e$, if $x_e \equiv 0 \pmod{7}$, then color edge $e$ white. If $x_e^2 \equiv 1 \pmod{7}$, color edge $e$ blue. If $x_e^2 \equiv 2 \pmod{7}$, color edge $e$ red. If $x_e^2 \equiv 4 \pmod{7}$, color edge $e$ green. It is not hard to see that this is a valid quasi-coloring.

To show that $c = \frac{1}{4}$ is indeed the smallest possible solution, we can use the graph shown in the previous solution.