Water flows through a horizontal tube of variable cross section....

Given volume flow rate, $Q = 1 \text{ cm}^3\text{/s} = 10^{-6} \text{ m}^3\text{/s}$

Cross-sectional area at $A$, $A_A = 4 \text{ mm}^2 = 4 \times 10^{-6} \text{ m}^2$

Cross-sectional area at $B$, $A_B = 2 \text{ mm}^2 = 2 \times 10^{-6} \text{ m}^2$

From the equation of continuity, the velocities at $A$ and $B$ are:

$v_A = \dfrac{Q}{A_A} = \dfrac{10^{-6}}{4 \times 10^{-6}} = 0.25 \text{ m/s}$

$v_B = \dfrac{Q}{A_B} = \dfrac{10^{-6}}{2 \times 10^{-6}} = 0.5 \text{ m/s}$

Using Bernoulli's equation for a horizontal tube:

$P_A + \dfrac{1}{2}\rho v_A^2 = P_B + \dfrac{1}{2}\rho v_B^2$

$P_A - P_B = \dfrac{1}{2}\rho(v_B^2 - v_A^2)$

Substituting the density of water $\rho = 1000 \text{ kg/m}^3$:

$P_A - P_B = \dfrac{1}{2} \times 1000 \times (0.5^2 - 0.25^2)$

$P_A - P_B = 500 \times (0.25 - 0.0625)$

$P_A - P_B = 500 \times 0.1875 = 93.75 \text{ N/m}^2$

Rounding to the nearest integer, we get $94 \text{ N/m}^2$.