Two identical bar magnets are dropped simultaneously from rest...

When a magnet falls through a copper tube, it induces eddy currents that create an upward magnetic drag force. The terminal velocity is reached when this upward drag force exactly balances the downward gravitational force ($mg$).

The induced EMF is proportional to the velocity $v$ of the magnet. The induced eddy current $I$ is given by $I = \dfrac{\text{EMF}}{R}$, where $R$ is the electrical resistance of the copper tube. Thus, the upward drag force $F_{drag} \propto I \propto \dfrac{v}{R}$.

At terminal velocity, $F_{drag} = mg$. Therefore, $\dfrac{v}{R}$ must be a constant for a given magnet and tube geometry. This implies that the terminal velocity $v$ is directly proportional to the resistance $R$ of the tube ($v \propto R$).

Copper is a metallic conductor, so its electrical resistance decreases as temperature decreases. Since Tube B is at a much lower temperature than Tube A, its resistance is lower ($R_B < R_A$).

Because $v \propto R$ and $R_B < R_A$, it follows that the terminal velocity in Tube B will be lower than in Tube A. Therefore, $v_A > v_B$.

Answer: $v_A > v_B$