strength of materials: normal stress of hollow circular tube

Because it is not stated in the problem, let us assume that the 48 kips is an axial load.

Solution
Inner radius, $r = 12.6 ~ \text{inches}$
Outer radius, $R = 12.6 + 0.5 = 13.1 ~ \text{inches}$
 

Calculate the cross-sectional area of the tube:
$A = \pi (R^2 - r^2)$
 

Calculate the normal stress.
$\sigma = \dfrac{P}{A}$ where $P = 48^k$
 

Your answer will be in the unit ksi (kilopound per square inch).