b

$ 18 = R_0 (1+ 20 \alpha )$

$ 20 = R_0 (1+ 50 \alpha )$

Dividing these two 

$\Rightarrow \frac{18}{20} = \frac{1+ 20 \alpha}{1+ 50 \alpha}$

$ \alpha = \frac{1}{250}$

$\text{Let Resistance become }21\Omega \text{ when temperature is t}$

$\Rightarrow 21 = R_0 (1+\alpha t)$ = R₀ (1+t/250)

$\text{Divide this to equation 1 }$

$\Rightarrow \frac{21}{18} = \frac{1+t/250}{1+20/250}$

$\Rightarrow t = 65^oC$

$\text{Rise in Temperature } = 65^oC - 15^oC= 50^oC$