There are $9$ marbles in the urn, so there are $\binom93$ different sets of $3$ marbles that you could draw without replacement. How many contain one marble of each color? To build such a set, you could pick either of the $2$ red marbles, any one of the $3$ white marbles, and any one of the $4$ blue marbles, so there are $2\cdot3\cdot4$ such sets. Each of the $\binom93$ sets is equally likely to be drawn, and $2\cdot3\cdot4$ of them are ‘successes’, so the probability of success is
If you draw with replacement, however, you can potentially draw the same marble twice, and you also have to take into account the order of the draws. On each of your $3$ draws you can get any of the $9$ marbles, so there are $9^3$ possible sequences of $3$ marbles that you can draw, and they’re all equally likely. How many of them contain one marble of each color? As in the first problem, there are $2\cdot3\cdot4=24$ different sets of $3$ marbles that will work, but each of them can be drawn in several different orders to give several different successful sequences of draws. In fact $3$ different objects can be arranged in $3!=6$ different orders, so each of those $24$ sets of $3$ marbles of $3$ different colors can be drawn in $6$ different orders. Thus, there are actually $6\cdot24$ successful sequences of $3$ draws. The final probability of success when you draw with replacement is therefore ... ?