Answer
2.7 cm.
Work Step by Step
Write Newton’s second law for the vertical direction. The maximum tension in the cable occurs when the acceleration is upward.
$$F_T-mg=ma$$
$$F_{T,max}=m(g+a)=35960\;N$$
With the safety factor of 8.0, the maximum tensile stress is 1/8th the tensile strength for steel, which is $500\times10^6\;N/m^2$.
The maximum stress occurs for the minimum area (minimum diameter).
$$\frac{F}{\pi (d/2)^2}=\frac{500\times10^6\;N/m^2}{8.0}$$
$$d=\sqrt{\frac{4(8.0)(35960\;N)}{\pi (500\times10^6\;N/m^2)}}=2.7\times10^{-2}\;m$$
