Answer
$$ d_{AB} \geq 22.57 \space mm$$
$$d_{BC} \geq 40.16 \space mm$$
Work Step by Step
1) Internal Forces:
$\bullet AB:$
Taking a cut between points A and B and applying the equilibrium equations:
$\sum {F_x} = 0; \space\space -60\space kN + N_{AB} = 0$
$N_{AB} = 60 \space kN \space (Tension)$
$\bullet BC:$
Taking a cut between points B and C and applying the equilibrium equations:
$\sum {F_x} = 0; \space\space -60\space kN + 2*125\space kN + N_{BC} = 0$
$N_{BC} = -190 \space kN \space (Compression)$
2)Minimum Cross-sectional Area/Diameter
From equation of the average normal stress:
$\sigma_{avg} = \frac{N_{int}}{A} \space\space \Rightarrow A = \frac{N_{int}}{\sigma_{avg}}$
Since $A = \frac{1}{4} \pi d^2 \Rightarrow \space\space d=\sqrt{\frac{4A}{\pi}}$
$\bullet AB:$
$A_{AB} \geq \frac{60*10^3\space kN}{150 * 10^6 \space Pa} = 4*10^{-4} \space m^2$
$d_{AB} \geq \sqrt{\frac{4A}{\pi}} = \boxed {22.57 \space mm} \space\space\space\leftarrow ANS1$
$\bullet BC:$
$A_{min, BC} \geq \frac{190*10^3\space kN}{150 * 10^6 \space Pa} = 1.267*10^{-3} \space m^2$
$d_{AB} \geq \sqrt{\frac{4A}{\pi}} = \boxed {40.16 \space mm} \space\space\space\leftarrow ANS2$
