 Solution

Assume an emitter at rest on the stellar surface and a static observer at infinity. Write $f(r)=1-2GM/(r{c}^2)$. The time-translation Killing energy of a photon is conserved, while the locally measured photon energy is proportional to $1/\sqrt{f(r)}$. ConsequentlyUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, $M=4\times 10^{30}\mathrm{kg}$ and $R=1.1\times 10^4\mathrm{m}$,The Schwarzschild radius is approximately $5.94\mathrm{km}$, below the given surface radius. The positive shift is therefore a gravitational redshift of an escaping photon, with received wavelength about $1.475$ times the emitted wavelength. A moving emitter would introduce an additional Doppler factor; that is outside this static-surface model. The weak-field estimate $z\simeq GM/(R{c}^2)=0.270$ is insufficient at this compactness. These stated constants are numerical conventions, not a claim about an unavailable historical handbook's last digits.