This problem is equivalent to finding a maximal independent set and colouring it white. To solve it, depth-first search on the graph, and colour the source node white.
If the graph is directed, change it to an undirected graph by replacing every edge with an undirected edge.
To clarify the terminology, "no two white colors are adjacent" means if 2 white vertices are connected by a directed edge, it violates this condition, so here the edge behaves as it is undirected. Since "between the white node and the black node there exists one directed path of length at most 2", so if there is an edge from the white node to the black node, it satisfies this criteria. Since "for each black node there exist at least one white node which is adjacent to it", if there is an edge from the black node to the white node, this criteria is also satisfied.
Let $x$ be the current node in the depth-first search.
If any of $x$'s neighbours has been coloured white, colour $x$ black. Otherwise, colour $x$ white.
Checking $x$'s neighbours can be done in $O(E)$ time for the entire graph over the depth first search, hence the entire algorithm is $O(V+E)$ time if the graph is stored as an adjacency list.
This algorithm terminates. We now show it works. If a node is coloured black, then some neighbour has to be white so that it can be coloured black.
Otherwise, a node is coloured white. The visited neighbours of the node (when the node is visited) are black. The subsequent unvisited neighbours of the nodes will not be coloured white as they are adjacent to a node that is coloured white. Hence all neighbours will be coloured black.
Hence, this satisfies your condition.
Best Answer
I have the answer for c = 1/2.
Just consider the white edges. As you might know, for every (undirected) graph you can can delete at most half of edges to get a bipartite graph. We'll call the partitions A and B. So there are at least half of the white edges between the group A and B. So if we color all the vertices of which ever group to white, there are at least half of the white edges which is connected to at least one white vertex. So we can choose to paint all the vertices of A to white, or all the vertices of B to white. (We'll choose it later)
Now that we partitioned the vertices into two groups, we want to see which partition should we paint to black, the other partition would be colored to white, and as mentioned, for whites the conditions is held.
The black edges between two partitions, will be connected to at least one black vertex whichever way. Now see the black edges inside the partitions. If black edges in partition A is more, we'll color all the vertices of the A to black. And if black edges in partition B is more, we'll do otherwise.
You can see for white and black the conditions is held.
Proof for the theorem used in this solution: Given a graph, partition it into 2 groups whichever way. If there was a vertex in one group, which more than half of its neighbors are in its group, move the vertex to the other partition. You can see this operation will end, and when ended, for each vertex at least half of its neighbors is in the other group. So at least half of the edges is in-between the partitions.
Update: In this proof, by
neighborswe mean the number of edges connected to one vertex.