= Solution
The given static induced surface charge and the dipole integral from part (b) give
$$
\mathbf p(t)=\alpha_E\mathbf E_{\rm inc}(t),\qquad
\alpha_E=4\pi\epsilon_0R^3\frac{\epsilon-1}{\epsilon+2}.
$$
Here $\epsilon$ is the dimensionless <relative permittivity>, as required by the stated $\epsilon\pm1,2$ expression; $\alpha_E$ is the SI <electric polarizability>. Take $\mathbf E_{\rm inc}(t)=E_0\cos\omega t\,\widehat{\mathbf e}$ and $k=\omega/c=2\pi/\lambda$. In the <Rayleigh scattering> regime the field is approximately uniform across the sphere. Its induced dipole radiates mean <power>
$$
\overline P_{\rm sca}=\frac{\mu_0\omega^4|\alpha_E|^2E_0^2}{12\pi c}.
$$
The incident time-averaged <light intensity> in <vacuum> is $I_{\rm inc}=\epsilon_0cE_0^2/2$. Dividing power by intensity gives the total <cross section>,
$$
\boxed{\sigma_{\rm sca}=\frac{k^4|\alpha_E|^2}{6\pi\epsilon_0^2}
=\frac{8\pi}3k^4R^6\left|\frac{\epsilon-1}{\epsilon+2}\right|^2
=\frac{128\pi^5R^6}{3\lambda^4}\left|\frac{\epsilon-1}{\epsilon+2}\right|^2.}
$$
The units are $\mathrm{m^2}$; the comparison is with the geometric projected area $\pi R^2$, not with a length:
$$
\boxed{\frac{\sigma_{\rm sca}}{\pi R^2}=\frac83\left(\frac{2\pi R}{\lambda}\right)^4
\left|\frac{\epsilon-1}{\epsilon+2}\right|^2\ll1.}
$$
The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered <power> has the characteristic $\lambda^{-4}$ dependence. Near a material resonance where $\epsilon+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.
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