\begin{aligned} &\Delta l=\frac{N \cdot l}{E \cdot A} \\ &l=1 \mathrm{~m} \\ &N=65 \cdot 10^{3} \mathrm{~N} \\ &\Delta l=1.2 \cdot 10^{-3} \mathrm{~m} \end{aligned}

We need to find the diameter of the cross-section, so the cross-section is circular, and therefore we calculate the area as:

\begin{aligned} A=\frac{\pi d^{2}}{4} \end{aligned}

We rearrange the formula and substitute the numerical data, calculate the diameter

\begin{aligned} &1.2 \cdot 10^{-3}=\frac{65 \cdot 10^{3} \cdot 1}{200 \cdot 10^{9} \cdot \frac{\pi d^{2}}{4}} \\ &1.2 \cdot 10^{-3}=\frac{260 \cdot 10^{3}}{200 \cdot 10^{9} \cdot \pi \cdot d^{2}} \quad \mid \cdot d^{2} \quad:\left(1.2 \cdot 10^{-3}\right) \\ &d^{2}=\frac{260 \cdot 10^{3}}{200 \cdot 10^{9} \cdot \pi \cdot 1.2 \cdot 10^{-3}} \\ &d=\sqrt{\frac{260 \cdot 10^{3}}{200 \cdot 10^{9} \cdot \pi \cdot 1.2 \cdot 10^{-3}}} \\ &d=0.0185 \mathrm{~m} \\ &d=1.85 \mathrm{~cm} \end{aligned}