The question is about arranging 6 balls (two identical red, two identical black, and two identical white) in such a way that the balls of the same color do not occupy any two consecutive cells.

Option 1 suggests 15 ways of arrangement, but this does not account for all permutations where no two consecutive cells contain balls of the same color.

Option 2 suggests 18 arrangements. This is the correct answer. The basis of this is there are 3! or 6 ways to arrange colors initially, but we have 2 balls of each color, so we multiply by 2^3 or 8. Results in 6*8=48 arrangements. However 2 consecutive balls of same colors can be arranged in 6 ways for each color and must be subtracted from initial 48 leaves 48-(6*3)=18

Option 3 suggests 24 arrangements, but this is more than the possible arrangements with the given condition.

Option 4 suggests 30 arrangements, but this also overestimates the possible arrangements that comply with the condition.

Therefore, Option 2 provides the correct answer.