Answer
According to you, the trip would take 27 years. According to an observer on Earth, it would take 68 years.
Work Step by Step
Solve for the speed from the length contraction relationship, equation 26–3a.
$$\mathcal{l} = \mathcal{l}_{o} \sqrt{1-\frac{ v^{2} } { c^{2} }} $$
$$v=c\sqrt{1-(\frac{\mathcal{l}}{\mathcal{l}_o})^2}$$
$$v=c\sqrt{1-(\frac{25ly}{62 ly})^2}$$
$$v=0.915c $$
Calculate the time from the speed and the contracted distance.
$$t=\frac{25ly}{0.915c}=27\;years$$
That is the elapsed time according to you. For an observer on Earth, the answer would be longer.
$$t_E=\frac{62ly}{0.915c}=68\;years$$
