= Solution
Assume an emitter at rest on the stellar surface and a static observer at infinity. Write $f(r)=1-2GM/(rc^2)$. The time-translation <Killing energy> of a <photon> is conserved, while the locally measured <photon> <energy> is proportional to $1/\sqrt{f(r)}$. Consequently
$$
\frac{\nu_\infty}{\nu_R}=\sqrt{f(R)},\qquad
\boxed{1+z=\frac1{\sqrt{1-2GM/(Rc^2)}}.}
$$
Using $G=6.67430\times10^{-11}\ {\rm m^3\,kg^{-1}\,s^{-2}}$, $c=299792458\ {\rm m\,s^{-1}}$, $M=4\times10^{30}\ {\rm kg}$ and $R=1.1\times10^4\ {\rm m}$,
$$
\frac{2GM}{Rc^2}=0.54008438,\qquad
\boxed{z=0.4745548\simeq0.475.}
$$
The <Schwarzschild radius> is approximately $5.94\ {\rm km}$, 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 $z\simeq GM/(Rc^2)=0.270$ is insufficient at this compactness. These stated constants are numerical conventions, not a claim about an unavailable historical handbook's last digits.
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