Answer
See explanation.
Work Step by Step
The spacecraft clock, like the internal clock of a moving muon, measures the proper time $\Delta t_o$. It will show the shorter elapsed time.
The value of $\gamma$ is $\frac{1}{\sqrt{1-u^{2}/c^{2}}}=1.000128$.
The Earth folks measure a dilated time interval of $\gamma \Delta t_o $.
The difference in elapsed time is $(\gamma-1)\Delta t_o=0.000128(365.25 days)(24hrs/day)=1.12 hours$.
