Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn.
Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$
Now, if we know a black ball is going to be removed in the second draw, then only one of the other two black balls, or one of the three red balls, could be removed from the second urn during the first draw.
So the probability of drawing a red ball in the first draw give that knowledge is:$$P(D_1=r \mid D_2=k) \\ = P(D_1=r \cap U_1=a \mid D_2=k)+P(D_1=r \cap U_1=b \mid D_2=k) \\ = P(U_1=a)P(D_1=r \mid U_1=a \cap D_2=k)+P(U_1=b)P(D_1=r \mid U_1=b \cap D_2=k) \\ = \frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{3}{5} \\ = \frac{7}{15}$$