We have the relation
`(1)/(eta)=e^(-nsigma)`
Here`(1)/(eta)=` attentution factor
`n=` no. of `Cd` nuclei per unit volume
`sigma=` effective cross section
`d=` thickness of the plate
Now `n=(rhoN_(A))/(M)`
`(rho=` density, `M=` Molar weight of `Cd, N_(A)=` Avogadro number.)
Thus `sigma=(M)/(rhoN_(A)d)In eta = 2.53kb`