Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 60 / 2 / i / Solution

The physical radius comes from the angular diameter distance, not the luminosity distance. In the specified flat cosmology,Using the supplied distance-integral approximation at $z=1$ and $c\simeq 3.00\times 10^5\mathrm{km}{\mathrm{s}}^{-1}$ gives $D_A\simeq (c/70)(0.8/2)=1714\mathrm{Mpc}$. The angular radius in radians is $(7/12)\pi /(180\times 3600)=2.828\times 10^{-6}$. ThusThis is a proper radius at the observation epoch; its comoving counterpart would be twice as large. The supplied approximate integral is sufficient here; integrating the specified Hubble parameter numerically gives a slightly smaller radius, about $4.82\mathrm{kpc}$.