A wire of length 1.20 m and diameter 0.50 mm is the unknown resistance in a meter bridge…

Step 1. The resistance. $X = 6.0\ \Omega \times \dfrac{36.0}{64.0} = 3.38\ \Omega$.

Step 2. The resistivity. $\rho = \dfrac{\pi d^2 X}{4L} = \dfrac{\pi \times (0.50 \times 10^{-3}\ \text{m})^2 \times 3.375\ \Omega}{4 \times 1.20\ \text{m}} = 5.52 \times 10^{-7}\ \Omega\,\text{m}$.

Step 3. The error. $\dfrac{\delta \rho}{\rho} = 2\,\dfrac{\delta d}{d} = 2 \times \dfrac{0.01}{0.50} = 0.040$, which is $4.0\ \%$.

What to learn. The diameter has the biggest effect on the error, so measure it with care, at several places along the wire.