Amir has identified why the book's answer and your's differ, Felix Brauer
The question you asked was:
An urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls. From the first urn are draws 2 balls and they dropped in the second urn. Then from the second urn is drawn one ball which is white. What is the probability that two balls transferred in second urn were black balls ?
The solution to this is, using Conditional Probability and the Law of Total Probability:
$$\begin{align}
\mathsf P(F_{BB}\mid S_W) & = \frac{
\mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB})
}{
\mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB})+\mathsf P(S_W\mid F_{BW})\,\mathsf P(F_{BW})+\mathsf P(S_W\mid F_{WW})\,\mathsf P(F_{WW})
}
\\
& = \frac{\frac 6 {12}{4\choose 2}/{7\choose 2}}{\left(\frac 6 {12}{4\choose 2}+\frac 7{12}{3\choose 1}{4\choose 1}+\frac 8{12}{3\choose 2}\right)/{7\choose 2}}
\\
& = \frac{6\cdot 6}{6\cdot 6+7\cdot 12+8\cdot 3}
\\ & = 1 / 4
\end{align}$$
As you, and Amir, obtained.
However if the question had asked:
Then from the second urn is drawn one ball which is black. What is the probability that two balls transferred in second urn were black balls ?
$$\begin{align}
\mathsf P(F_{BB}\mid S_B) & = \frac{
\mathsf P(S_B\mid F_{BB})\,\mathsf P(F_{BB})
}{
\mathsf P(S_B\mid F_{BB})\,\mathsf P(F_{BB})+\mathsf P(S_B\mid F_{BW})\,\mathsf P(F_{BW})+\mathsf P(S_B\mid F_{WW})\,\mathsf P(F_{WW})
}
\\
& = \frac{\frac 6 {12}{4\choose 2}/{7\choose 2}}{\left(\frac 6 {12}{4\choose 2}+\frac 5{12}{3\choose 1}{4\choose 1}+\frac 4{12}{3\choose 2}\right)/{7\choose 2}}
\\
& = \frac{6\cdot 6}{6\cdot 6+5\cdot 12+4\cdot 3}
\\ & = 1 / 3
\end{align}$$
As you say the book gives.