Answer
$5.0\times10^{-3}\,(C^{\circ})^{-1}$
Work Step by Step
$R=R_{0}(1+\alpha\Delta T)$ where $\alpha$ is the temperature coefficient of resistivity.
$\implies \alpha=\frac{R-R_{0}}{R_{0}\Delta T}=\frac{43.7\,\Omega-38.0\,\Omega}{38.0\,\Omega\times(55-25)^{\circ}C}=5.0\times10^{-3}\,(C^{\circ})^{-1}$
