A bag $B$ contains $4$ white balls and $2$ black balls.

Probability of selecting $1$ white ball from bag $B$$=P_1=\frac{4}{6}$

Another bag $C$ contains $3$ white balls and $5$ black balls

Probability of selecting $1$ white ball from bag $C$$=P_2=\frac{3}{8}$

Hence, the probability that the two balls drawn are white$=P_1\times P_2=\frac{4}{6}\times \frac{3}{8}=\frac{1}{4}$