Answer
$\eta=3.5\times10^{-3}Pa\cdot s$
Work Step by Step
Find $\rho_b$ on page 261
10-3b(page 263): $\Delta P=\rho g\Delta h$
Poiseuille's equation 10-9(page 280): $Q=\frac{\pi\Delta Pr^4}{8\eta l}$
$P_2-P_1=\rho_b gh=(1.05\times10^3\frac{kg}{m^3})(9.8\frac{m}{s^2})(1.40m)=1.44\times10^4\frac{N}{m^2}$
$\eta=\frac{\pi\Delta Pr^4}{8Q l}=\frac{\pi(1.44\times10^4\frac{N}{m^2})(0.20\times10^{-3}m)^4}{8(4.1\frac{cm^3}{min}\times\frac{1 min}{60 s}\times\frac{1m^3}{10^6m^3})(0.038m)}$
$=3.5\times10^{-3}Pa\cdot s$