Yes. There are nine ways for there to be an even number on cube one and an odd number on cube two. The conditional probability is the same for both events.
Work Step by Step
number cubes are numbered $1, 2, 3, 4, 5,$ and $6$ (three even numbers and three odd numbers) P(even on cube 1 | odd on cube 2) = $1/4$ P(odd on cube 2 | even on cube 1) = $1/4$