$R_{t}=R_{0}(1+\alpha t)$

where $R_{t}$ is the resistance of the wire at $t\,^{o} \mathrm{C}$

$R_{0}$ is the resistance of the wire at $0\,^{o} \mathrm{C}$

and $\alpha$ is the temperature coefficient of resistance.

$\Rightarrow R_{50}=R_{0}(1+50 \alpha)$         ......$(i)$

$R_{100}=R_{0}(1+100 \alpha)$       .......$(ii)$

From $(i),$ $R_{50}-R_{0}=50 \alpha R_{0}$        .....$(iii)$

From $(ii),$ $R_{100}-R_{0}=100 \alpha R_{0}$        ....$(iv)$

Dividing $(iii)$ by $(iv),$ we get

$\frac{R_{50}-R_{0}}{R_{100}-R_{0}}=\frac{1}{2}$

Here, $R_{50}=5\, \Omega$ and $R_{100}=6\, \Omega$

$\therefore  $   $\frac{5-R_{0}}{6-R_{0}}=\frac{1}{2}$

or, $6-R_{0}=10-2 R_{0}$ or, $R_{0}=4\, \Omega$