Given: ${\mathrm{R}}_{50}=5 \mathrm{\Omega }, {\mathrm{R}}_{100}=6 \mathrm{\Omega }$

${\text{R}}_{\text{t}}={\text{R}}_0\left(1+\alpha \text{t}\right)$

Where ${\mathrm{R}}_{\mathrm{t}}$ = resistance of a wire at $\mathrm{t} °\mathrm{C}$, ${\mathrm{R}}_0$ = resistance of a wire at $0 °\mathrm{C}$, $\alpha$= temperature coefficient of resistance.

$\therefore$  ${\text{R}}_{50}={\text{R}}_0\left[1+\alpha 50\right]$

and   ${\text{R}}_{100}={\text{R}}_0\left[1+\alpha 100\right]$

or      ${\text{R}}_{50}-{\text{R}}_0={\text{R}}_0\alpha \left(50\right)$                         

${\text{R}}_{100}-{\text{R}}_0={\text{R}}_0\alpha \left(100 \right)$                  

Dividing (i) by (ii), we get

$\frac{5-{\text{R}}_0}{6-{\text{R}}_0}=\frac{1}{2}$

or      $10-2{\mathrm{R}}_0=6-{\mathrm{R}}_0$

or       ${\text{R}}_0=4 \mathrm{\Omega }$