Resistance of the wire is measured as 2   Ω and 3   Ω at 10 ° C and 30 ° C…

Resistance of a material is related to temperature by expression $R=R_0\left(1+\alpha \Delta T\right)$

Here, at $10°\mathrm{C}$, we have  $2=R_0\left(1+10\alpha \right) ...\left(i\right)$

And at $30°\mathrm{C}$, we have $3=R_0\left(1+30\alpha \right) ...\left(ii\right)$

From equation $\left(i\right)$, $R_0=\frac{2}{\left(1+10\alpha \right)}$

Putting this value in equation $\left(ii\right)$, 

$$\begin{gathered} 3=\frac{2}{1+10\alpha }\left(1+30\alpha \right) \\ \Rightarrow 3+30\alpha =2+60\alpha \end{gathered}$$

$\Rightarrow 1=30\alpha$

Hence, temperature coefficient of the material of wire is $\alpha =\frac{1}{30}=0.033°{\mathrm{C}}^{-1}$