= Solution
In the stipulated constant-speed model, the Earth-frame <speed> is $v=D/t=3.844\times10^8/(76\times3600)\simeq1.405\times10^3$ m/s. The Earth and Moon are stationary endpoints in this model, so their Earth-frame separation $D$ is the <proper length>. The spacecraft-frame separation, measured simultaneously in that frame, is $D'=D/\gamma=D\sqrt{1-v^2/c^2}$. Thus
$$
D-D'=\frac{D(v/c)^2}{1+\sqrt{1-(v/c)^2}}\simeq\frac{Dv^2}{2c^2}\simeq4.22\times10^{-3}\,\mathrm m.
$$
\b[The spacecraft-frame <distance> is shorter by about 4.2 mm.] This compares simultaneous endpoint measurements in two frames, not the coordinate difference between a launch event and an arrival event. The <small-speed length contraction estimate> also avoids numerical cancellation when evaluating the tiny contraction.
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