= Solution
Use the <inertial frame> in which the departure point and destination are stationary, with separation $d$. The astronauts' elapsed <proper time> is $\tau=t/\gamma$, whereas $d=vt$. Hence
$$
\gamma\beta=\frac d{c\tau}=\frac45,\qquad
\gamma^2-(\gamma\beta)^2=1,
\qquad \gamma=\frac{\sqrt{41}}5,\qquad \beta=\frac4{\sqrt{41}}.
$$
The Earth-frame <travel time> is $t=\sqrt{41}\,\mathrm{yr}\simeq6.40\,\mathrm{yr}$ and the <speed> is approximately $0.625c$. The spacecraft's <relativistic kinetic energy> is therefore
$$
\boxed{K=(\gamma-1)mc^2
=\left(\frac{\sqrt{41}}5-1\right)(10^5\,\mathrm{kg})c^2
\simeq2.52\times10^{21}\,\mathrm J.}
$$
The five years are not the time to use in $v=d/t$ in the departure frame. This is the cruising <kinetic energy>, with departure and arrival <acceleration> intervals omitted, rather than a complete propulsion <energy> budget.
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