The possible outcomes are $X=0,X=1$, and $X=2$. In order to get no white marbles ($X=0$), you must draw two black marbles. There are $\binom82$ pairs of marbles altogether, and there are $\binom32$ pairs of black marbles, so the probability of drawing a black pair is
$\frac{\binom32}{\binom82}=\frac3{28}\;.$
That is, $\Bbb P(X=0)=\frac3{28}$.
There are $\binom52$ pairs of white marbles, so the probability of drawing one of those pairs is
$\frac{\binom52}{\binom82}=\frac{10}{28}=\frac5{14}\;.$
That is, $\Bbb P(X=2)=\frac5{14}$.
The total probability of all three possible outcomes must be $1$, so we must have
$\Bbb P(X=1)=1-\frac3{28}-\frac5{14}=1-\frac{13}{28}=\frac{15}{28}\;,$
and we have our probability distribution:
$\begin{align*} \Bbb P(X=0)&=\frac3{28}\\ \Bbb P(X=1)&=\frac{15}{28}\\ \Bbb P(X=2)&=\frac5{14}\;. \end{align*}$
We can double-check this by calculating $\Bbb P(X=1)$ directly. Each of the $5$ white balls can be paired with any of the $3$ black balls to make a pair with one white ball, so there are $15$ such pairs, and as we’ve seen already, there are $\binom82=28$ pairs altogether, so the probability of drawing a pair with one ball of each color is indeed $\frac{15}{28}$.