Chapter 15 - Fluids - Problems and Conceptual Exercises - Page 535: 85

$95.3KPa$

We know that $A=\pi(\frac{d}{2})^2$ $A=\pi(\frac{0.0139}{2})^2=1.517\times 10^{-4}m^2$ We can find the required pressure as $P=\frac{F}{A}$ We plug in the known values to obtain: $P=\frac{3.25lb}{1.517\times 10^{-4}}(\frac{1N}{0.225lb)}=95.3KPa$