Answer
$\frac{d\rho}{dT} = 8 \times 10^{-11} \Omega . m /K$
Work Step by Step
For copper,
$\frac{d\rho}{dT} = [\rho\alpha]_{Cu} $
Where
$\rho = 2 \times 10^{-8} \Omega . m$
$\alpha = 4 \times 10^{-3} K^{-1} $
Substitute into equation
$\frac{d\rho}{dT} = (2 \times 10^{-8} \Omega . m)(4 \times 10^{-3} K^{-1} )$
$\frac{d\rho}{dT} = 8 \times 10^{-11} \Omega . m /K$
