Let $U_i$ be the $i$th urn just like you described, and let $A =\{\text{Picked a white ball}\},B_i = \{\text{Pick $U_i$}\}.$ Then the event "all 3 balls in the selected urn are white" is the same as event $B_4$.
Thus, if we use conditioning, the probability we are interested in is
\begin{align*}
P(B_4|A)&=\frac{P(B_4A)}{P(A)}\\
&=\frac{P(A|B_4)P(B_4)}{P(A)}\\
&=\frac{P(A|B_4)P(B_4)}{P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+P(A|B_3)P(B_3)+P(A|B_4)P(B_4)}\\
&=\frac{1\cdot \frac{1}{4}}{0(1/4)+(1/3)(1/4)+(2/3)(1/4)+1(1/4)}\\
&=\frac{1}{2}.
\end{align*}
Method I: (Bayes) There are two scenarios in which a red ball is observed:
I: A white ball is transferred (probability $\frac 35$). In this case, a red ball is observed with probability $\frac 12$. Thus this scenario has probability $$\frac 35 \times \frac 12 = \frac 3{10}$$
II: A red ball is transferred (probability $\frac 25$). Now the probability of drawing a red ball is $\frac 34$ Thus this scenario has probability $$\frac 25\times \frac 34 =\frac 3{10}$$
We see that the two scenarios contribute equally, thus the probability that it was a white ball that was transferred initially is $\boxed {\frac 12}$
Note: as our prior was that the probability the transferred was white was $\frac 35$ we see that the observation of the red ball has caused us to lower our estimate for the probability
Method II (Conditional Probability). A priori, the universe here consists of four possible events: $(W,W),(W,R),(R,R),(R,W)$ according to the color of the transferred ball and the color of the drawn ball. A routine calculation shows that $(W,W),(W,R),(R,R)$ each have probability $\frac 3{10}$ and $(R,W)$ has probability $\frac 1{10}$. We are asked for the probability that the transferred ball is $W$ conditioned on the fact that the drawn ball is $R$ and inspection now shows the answer to be $\frac 12$.
Best Answer
Let A be the event that urn A chosen, B be the event that Urn B chosen,
and W be the event that a white ball is chosen.
Using Bayes' Rule and the law of total probability,
P(From Urn A | it is white) $= \dfrac{P(A)\cdot P(W|A)}{P(A)\cdot P(W|A) + P(B)\cdot(P(W|B)} =\dfrac{\dfrac12 \cdot\dfrac23}{\dfrac12\cdot\dfrac23+\dfrac12\cdot\dfrac13}=\dfrac23$