= Solution
The prescribed surface <charge density> is that of a <uniformly polarized sphere> and has zero total charge because $\int\cos\theta\,d\Omega=0$. Its <electric dipole moment> at $t=0$ is
$$
\mathbf p_0=\int_{r=R}\mathbf r\,\sigma\,dA
=\sigma_0R^3\int\widehat{\mathbf r}\cos\theta\,d\Omega.
$$
The transverse components cancel by <axisymmetry>, while the $z$ component is
$$
p_{0,z}=2\pi\sigma_0R^3\int_0^\pi\cos^2\theta\sin\theta\,d\theta
=\frac{4\pi}3\sigma_0R^3.
$$
Thus
$$
\boxed{\mathbf p_0=p_0\widehat{\mathbf z},\qquad p_0=\frac{4\pi}3\sigma_0R^3.}
$$
The angular surface distribution is a pure degree-one <spherical harmonic>. Hence its static external <electric potential> is exactly that of this dipole, with no higher electric <multipole moments>; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.
For right-handed uniform rotation about $+\widehat{\mathbf x}$, the rigidly attached dipole is
$$
\mathbf p(t)=p_0\left(-\widehat{\mathbf y}\sin\omega t+\widehat{\mathbf z}\cos\omega t\right),\qquad
\ddot{\mathbf p}=-\omega^2\mathbf p.
$$
Consequently $|\ddot{\mathbf p}|^2=p_0^2\omega^4$ is constant. The <electric dipole radiation> formula gives
$$
\boxed{\overline P=\frac{\mu_0p_0^2}{6\pi c}\omega^4
=\frac{8\pi\mu_0\sigma_0^2R^6}{27c}\,\omega^4\equiv K\omega^4.}
$$
There is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading <electric-dipole approximation for radiation>, requiring $\omega R/c\ll1$ and observation in the <radiation zone>. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The <relative permittivity> enters the induced-polarization problem in part (d), not this prescribed total dipole moment.
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