We have, $R_t=R_0 \left(1+\alpha t\right)$

Here, $20=R_0 \left(1+20\alpha \right)$

$60=R_0 \left(1+500\alpha \right)$

$\Rightarrow \frac{60}{20}=\frac{R_0 \left(1+500\alpha \right)}{R_0 \left(1+20\alpha \right)}$

$3=\frac{1+500 \alpha }{1+20 \alpha }$

$\Rightarrow 3 \left(1+20 \alpha \right)=1+500 \alpha$

$\Rightarrow 3+60\alpha =1+500\alpha$

$\Rightarrow 440 \alpha =2$

$\Rightarrow \alpha =\frac{2}{440}=\frac{1}{220}I°\mathrm{C}$

Also given $R_t=25 °\mathrm{C}, t= ?$

Now, $\frac{20}{25}=\frac{R_0 \left(1+\frac{20}{220}\right)}{R_0 \left(1+\frac{t}{220}\right)}$

$4 \left(1+\frac{t}{220}\right)=5 \left(1+\frac{20}{220}\right)$

On solving, 

$t=80 °\mathrm{C}$