Answer
$$d_{IN} = 14.93 \space mm$$
Work Step by Step
1) Internal Forces:
Taking a cut through point C and applying equilibrium equations:
$\sum{F_y} = 0; \space\space N_{int} - 1200 \space N = 0 \space \Rightarrow N_{int} = 1200 \space N$
2) Cross sectional Area:
Cross sectional area of an annular cylinder:
$A_{SXN} = \frac{\pi}{4}(d_{OUT}^2 - d_{IN}^2)$
3) Average Normal Stress:
$\sigma_{AVG} = \frac{N_{INT}}{A_{SXN}}$
Solving for $A_{SXN}$:
$A_{SXN} = \frac{\pi}{4}(d_{OUT}^2 - d_{IN}^2) = \frac{N_{INT}}{\sigma_{AVG}}$
Solving the above equation for $d_{IN}$
$d_{IN} = \sqrt{d_{OUT}^2 - \frac{N_{INT}}{\sigma_{AVG}} * \frac{4}{\pi}} = \sqrt{(25*10^{-3} \space m)^2 - \frac{1200 \space N}{3.80 *10^6 \space Pa} * \frac{4}{\pi}}$
$d_{IN} = \boxed{14.93 \space mm} \space\space \leftarrow ANS$
