d
variation of resistance with temperature
$16=R_{0}\left[1+\alpha\left(15-T_{0}\right)\right]$
$20=R_{0}\left[1+\alpha\left(100-T_{0}\right)\right]$
Assuming ${T}_{0}=0^{\circ} {C}$, as a general convention.
$\Rightarrow \frac{16}{20}=\frac{1+\alpha \times 15}{1+\alpha \times 100} \Rightarrow \alpha=0.003^{\circ} {C}^{-1}$