KEY CONCEPT : We know that

$${R_t} = R{}_0\left( {1 + \alpha t} \right),$$

where $${R_t}$$ is the resistance of the wire at $${t^ \circ }C,$$

$${R_0}$$ is the resistance of the wire at $${0^ \circ }C$$

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

$$ \Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$${R_{100}} = {R_0}\left( {1 + 100\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$\left( i \right),\,{R_{50}} - {R_0} = 50\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$

From $$\left( {ii} \right),{R_{100}} - {R_0} = 100\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iv} \right)$$

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

$${{{R_{50}} - {R_0}} \over {{R_{100}} - R{}_0}} = {1 \over 2}$$

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

$$\therefore$$ $${{5 - {R_0}} \over {6 - {R_0}}} = {1 \over 2}$$

or, $$6 - {R_0} = 10 - 2{R_0}$$

or, $${R_0} = 4\Omega .$$