Player B has a winning strategy. One way to see this is to divide the 3-digit numbers into 8 groups with the following property: For every two numbers $a$, $b$ from the same group, $a > b$, one has $\frac{a}{b} < \frac{4}{3}$.

$$\begin{align*} G_1 &= \{100, \dots, 133\}, \\ G_2 &= \{134, \dots, 178\}, \\ G_3 &= \{179, \dots, 238\}, \\ G_4 &= \{239, \dots, 318\}, \\ G_5 &= \{319, \dots, 425\}, \\ G_6 &= \{426, \dots, 567\}, \\ G_7 &= \{568, \dots, 757\}, \\ G_8 &= \{758, \dots, 999\}. \end{align*}$$

For a justification it suffices to note that the ratio of the last number and the first number in a group is less than $\frac{4}{3}$, for instance, $\frac{567}{426} < \frac{4}{3}$.

Since there are 13 different numbers chosen by A and $13 > 8$, some two of them are in the same group $G_i$, $1 < i < 8$. Let them be $a$ and $b$ with $a > b$; then $1 < \frac{a}{b} < \frac{4}{3}$. Among the remaining 11 numbers B can find another two with the same property, say $c$ and $d$ with $1 < \frac{c}{d} < \frac{4}{3}$. This is because $11 > 8$. Finally B can select one more analogous pair $e, f$, satisfying $1 < \frac{e}{f} < \frac{4}{3}$, because there are still $9 > 8$ numbers left. Now adding up gives $3 = 1+1+1 < \frac{a}{b} + \frac{c}{d} + \frac{e}{f} < 3 \cdot \frac{4}{3} = 4$, and the task of B is complete.