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

Reason : Kirchoff 's loop rule follows from conservation of momentum.

$[1]$ battery emf is increased
$[2]$ all resistances are increased by $10\,ohms$
$[3]$ all resistances are made five times
$[4]$ the battery and the galvanometer are interchanged

Statement $1 :$ The possibility of an electric bulb fusing is higher at the time of switching $ON.$
Statement $2:$ Resistance of an electric bulb when it is not lit up is much smaller than when it is lit up.