Expansion redshifts <photon> energies and preserves a <blackbody spectrum>, so a <comoving observer> at $z=2$ sees $T=T_0(1+z)\simeq8.18$ K. A one-sided dipole excess of 0.105% means $v/c=0.00105$, hence $\boxed{v\simeq315\ \mathrm{km\,s^{-1}}}$. Motion is toward the hotter region. The peak-to-peak contrast is twice this, so using 0.210% for $v/c$ would double the inferred <speed> incorrectly.

Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S2 / 3 / 1 / Solution

 Solution (source code)