Bayes' Theorem is one way to do this, but it isn't necessary.
Suppose the balls were labeled $W_1,W_2,B_1,B_2$ and that the second draw was $W_1$. Then we know the first ball must have been one of $W_2,B_1,B_2$. Nothing distinguishes these except for color so any one of them is equally likely to have been the first one chosen. Thus the probability that the first draw was Black is $\frac 23$ and the probability that it was White is $\frac 13$.
To help with intuition, here is another way to look at it. Imagine we draw out all $4$ balls (though in the end we only care about the first two draws). There are $\binom 42=6$ equally probable ways the balls might be ordered. We have $$WWBB,WBWB,WBBW,BBWW,BWBW,BWWB$$ Of these, only three have $W$ as the second draw, namely: $$WWBB,BWBW,BWWB$$
Thus, we know we are in one of these three states, each with probability $\frac 13$. We remark that in two of these three cases, the first ball is $B$.
To stress: I am assuming here that the first draw is NOT returned to the bag. It is drawn and then hidden away somewhere, not in the bag. That is critical. If, to the contrary, you show it to a friend (who notes the color) and then return it to the bag, then the two draws are entirely independent. In the context of my solution above, if you have replaced the first draw in the bag then there was no reason to exclude $W_1$ from the list of possibilities.
If a bag contains 5 Black marbles and 4 White marbles, and you withdraw 5 marbles at random without replacement, then then number $X$ of White marbles among the 5 withdrawn has a hypergeometric distribution with
$$P(X = k) = \frac{{4 \choose k}{5 \choose 5-k}}{{9 \choose 5}},$$
for $k =0, 1, 2,3,4.$
In particular, $$P(X = 3) =
\frac{{4 \choose 3}{5 \choose 2}}{9\choose 5}
= \frac {4\cdot10}{126} = 0.3174603.$$
In R statistical software, where dhyper is a hypergeometric PDF, you can make a probability table for the distribution as shown below. [You can ignore row numbers in brackets.]
k = 0:5; PDF = dhyper(k, 4,5, 5)
cbind(k, PDF)
k PDF
[1,] 0 0.007936508
[2,] 1 0.158730159
[3,] 2 0.476190476
[4,] 3 0.317460317 # Shown above
[5,] 4 0.039682540
[6,] 5 0.000000000 # Impossible to get 5 White
Notice that I tried to find the probability of getting 5 white marbles, which is impossible because there are only 4 white marbles in the bag. In writing the PDF you can either (i) be careful to restrict $k$ only to possible values or (ii) use the convention that the binomial coefficient ${a \choose b} = 0,$ if integer $b$ exceeds integer $a.$ If your text includes a formal statement of the hypergeometric PDF, you should notice which method is used.
Here is a plot of the specific hypergeometric distribution mentioned in your problem.
Computations of $\mu = E(X) = \sum_{k=0}^5 k*p(k) = 5(4/9) = 2.2222$
and
$$\sigma^2 = Var(X) = \sum_{k=0}^5 (k-\mu)^2p(k)\\
= \sum_{k=0}^5 k^2p(k) - \mu^2 = 0.6173$$ are shown below.
You may find formulas for these in your text.
mu = sum(k*PDF); mu
[1] 2.222222
vr = sum((k-mu)^2*PDF); vr
[1] 0.617284
sum(k^2*PDF) - mu^2
[1] 0.617284
Best Answer
Let $W_k$ be the event of drawing a white marble on the $k$-th draw, and $B_k$ the complementary event.
So your solution should look somewhat like:
(a) $$\begin{align}\mathsf P(W_1\mid W_2)&=\dfrac{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)}{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)+\mathsf P(B_1)~\mathsf P(W_2\mid B_1)}\\[1ex]&=\dfrac{\tfrac{4}{4+9}\tfrac{3}{3+11}}{\tfrac{4}{4+9}\tfrac{3}{3+11}+\tfrac{9}{4+9}\tfrac{6}{6+8}}&=\dfrac{4(4-1)}{4(4-1)+9(4+2)}\\[1ex]&=\dfrac{2}{11}&\checkmark\end{align}$$
The others use Bayes' Rule too. Hint: tackle (c) first.
(b) $$\begin{align}\mathsf P(W_1\mid W_3)&=\dfrac{\mathsf P(W_1,W_2,W_3)+\mathsf P(W_1,B_2,W_3)}{\mathsf P(W_1,W_2,W_3)+\mathsf P(W_1,B_2,W_3)+\mathsf P(B_1,W_2,W_3)+\mathsf P(B_1,B_2,W_3)}\\[2ex]&=\tfrac{\mathsf P(W_1)~\big(\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)+\mathsf P(B_2\mid W_1)~\mathsf P(W_3\mid W_1,B_2)\big)}{{\mathsf P(W_1)~\big(\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)} + {\mathsf P(B_2\mid W_1)~\mathsf P(W_3\mid W_1,B_2)\big)+\mathsf P(B_1)~\big(\mathsf P(W_2\mid B_1)~\mathsf P(W_3\mid B_1,W_2)+\mathsf P(B_2\mid B_1)~\mathsf P(W_3\mid B_1,B_2)\big)}}\end{align}$$
(c) $$\begin{align}\mathsf P(W_1\mid W_2,W_3)&=\dfrac{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)}{\mathsf P(W_1)~\mathsf P(W_2\mid W_1)~\mathsf P(W_3\mid W_1,W_2)+\mathsf P(B_1)~\mathsf P(W_2\mid B_1)~\mathsf P(W_3\mid B_1,W_2)}\end{align}$$