 Solution

Expansion redshifts photon energies and preserves a blackbody spectrum, so a comoving observer at $z=2$ sees $T=T_0(1+z)\simeq 8.18$ K. A one-sided dipole excess of 0.105% means $v/c=0.00105$, hence $\boxed{v\simeq 315\mathrm{km}{\mathrm{s}}^{-1}}$. 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.