Given here:
Mass density, ρ=ρ0(1−L2x2)kgm−1
Let us suppose there is an element of length dx at a distance x from left must end.

Mass of this element is dm=ρdx=ρ0(1−L2x2)
The centre of mass of this element is Xcm=∫dm∫xdm
=∫0lρ0(1−L2x2)dxρ0∫0l(x−L2x2)xdx
=L−3L2L32L2−4L2L4=32L4L2=83L
Hence, the value of α=8.