= Solution
The physical radius comes from the <angular diameter distance>, not the luminosity distance. In the specified flat cosmology,
$$
D_A(z)=\frac{c}{1+z}\int_0^z\frac{dz'}{H(z')}.
$$
Using the supplied distance-integral approximation at $z=1$ and $c\simeq3.00\times10^5\,\mathrm{km\,s^{-1}}$ gives $D_A\simeq(c/70)(0.8/2)=1714\,\mathrm{Mpc}$. The angular radius in radians is $(7/12)\pi/(180\times3600)=2.828\times10^{-6}$. Thus
$$
\boxed{R_d=D_A\theta\simeq4.85\,\mathrm{kpc}.}
$$
This 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 $4.82\,\mathrm{kpc}$.
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