A tube with cross sections at A and B , having areas of 4 mm^2 and...

Given the volume flow rate of water entering through cross-section $B$ is $Q = 1 \text{ cm}^3 \text{ s}^{-1}$.

The area of cross-section at $B$ is given as $A_B = 2 \text{ mm}^2$.

Converting the area into $\text{cm}^2$:
$A_B = 2 \times 10^{-2} \text{ cm}^2 = 0.02 \text{ cm}^2$

The volume flow rate is the product of the cross-sectional area and the speed of the fluid. Therefore, the speed of water at $B$, $v_B$, can be calculated using the relation:
$Q = A_B v_B$

Substituting the given values:
$1 \text{ cm}^3 \text{ s}^{-1} = 0.02 \text{ cm}^2 \times v_B$

$v_B = \dfrac{1}{0.02} \text{ cm s}^{-1} = 50 \text{ cm s}^{-1}$

The other given parameters (area at $A$, distance between $A$ and $B$, and acceleration due to gravity) are extra information and are not required to find the speed at $B$ since the flow rate is explicitly provided.

Answer: $50 \text{ cm/s}$