Chapter 15 - Fluids - Problems and Conceptual Exercises - Page 533: 58

$2.5\times 10^7$

We know that $n=\frac{A_aV_a}{A_cv_c}$ $\implies n=\frac{\frac{\pi}{4}d_a^2V_a}{\frac{\pi}{4}d_c^2v_c}$ $n=(\frac{d_a}{d_c})^2\frac{v_a}{v_c}$ We plug in the known values to obtain: $n=(\frac{0.0050}{1.0\times 1.0\times 10^{-5}})^2(\frac{1.0}{0.01})=2.5\times 10^7$