Let us consider a elemental length ‘d x’ at ‘x’ distance from one end as shown.

Let ‘dm’ be the mass of elemental length 

$$\begin{gathered} dm=\lambda .dx={\lambda }_0(1-\frac{x^2}{l^2})dx \\ {x}_{cm}=\frac{\int xdm}{\int d{m}_l}=\frac{{\lambda }_0{\displaystyle \underset{0}{\overset{l}{\int }}(x-\frac{x^3}{l^2})}dx}{{\displaystyle \underset{0}{\overset{l}{\int }}{\lambda }_0(1-\frac{x^2}{l^2})dx}}=\frac{\frac{l^2}{2}-\frac{l^4}{4l^2}}{l-\frac{l^3}{3l^2}}=\frac{3l}{8} \end{gathered}$$

Comparing it with $x_{cm}=\frac{3L}{\alpha }m$, we get  $\alpha =8$