The resistivity of a wire is ρ = (Rπ d²)/4l…

Step 1. The resistivity. $\rho = \dfrac{R\pi d^2}{4l}$ $= \dfrac{4.0\ \Omega \times \pi}{4 \times 0.800\ \text{m}}$ $\times (0.50 \times 10^{-3}\ \text{m})^2$ $= 9.8 \times 10^{-7}\ \Omega\,\text{m}$.

Step 2. The three terms. $\dfrac{\Delta R}{R} = \dfrac{0.1}{4.0} = 0.025$. $2\dfrac{\Delta d}{d} = 2 \times \dfrac{0.01}{0.50} = 0.040$. $\dfrac{\Delta l}{l} = \dfrac{0.1}{80.0} = 0.00125$.

Step 3. The sum. $\dfrac{\Delta \rho}{\rho}$ $= 0.025 + 0.040 + 0.00125$ $= 0.066$, which is $6.6\%$.

Which to improve. The diameter gives $0.040$ out of $0.066$, the largest part. A screw gauge with a smaller least count, or many readings of the diameter, would help most.