Answer
1.3 cm.
Work Step by Step
With the safety factor of 7.0, the maximum shear stress is 1/7th the shear strength for iron, which is $170\times10^6\;N/m^2$.
The maximum stress occurs for the minimum area (minimum diameter).
$$\frac{F}{\pi (d/2)^2}=\frac{170\times10^6\;N/m^2}{7.0}$$
$$d=\sqrt{\frac{4(7.0)(3300\;N)}{\pi (170\times10^6\;N/m^2)}}=1.3\times10^{-2}\;m$$
