Answer
See the detailed answer below.
Work Step by Step
Let's assume that the Earth's frame is $\rm S$ and the rocket's frame is $\rm S'$.
$$\color{blue}{\bf [a]}$$
Your round trip distance is twice the distance between the Earth and the Betelgeuse star, so $d=2L=2\times 430\;\rm ly=\bf 860\;\rm ly$.
You need to measure a proper time of 20 years from your frame.
So, $\tau=20\;\rm y$, where
$$\tau= \sqrt{1-\frac{v^2}{c^2}}\;\cdot t\tag 1$$
Your needed speed relative to the Earth is
$$v=\dfrac{2L}{t}\tag 2$$
Solving (1) for $t$ and plug into (2),
$$v=\dfrac{2L}{\tau} \sqrt{1-\frac{v^2}{c^2}} $$
$$\dfrac{\tau}{2L} v= \sqrt{1-\frac{v^2}{c^2}} $$
Squaring both sides;
$$\dfrac{\tau^2}{4L^2}v^2= 1-\frac{v^2}{c^2} $$
$$\dfrac{\tau^2}{4L^2}v^2+\frac{v^2}{c^2} = 1$$
$$v^2\left[ \dfrac{\tau^2c^2+4L^2}{4L^2c^2} \right] = 1$$
$$v=\sqrt{ \dfrac{4L^2c^2}{\tau^2c^2+4L^2} }$$
$$v=\sqrt{ \dfrac{4L^2 }{\tau^2c^2+4L^2} }\;\cdot c$$
Plug the known; where $c=\rm 1\;ly/y$
$$v=\sqrt{\rm \dfrac{4(430\;ly)^2 }{(20\;y)^2\left(\frac{1\;ly}{\;y}\right)^2 +4(430\;ly)^2} }\;\cdot c$$
$$\boxed{v=\color{red}{\bf 0.99973} c}$$
$$\color{blue}{\bf [b]}$$
The energy needed to accelerate your rocket is given by
$$\Delta E=E_f-E_0$$
where $E_0=mc^2$ which is the rest energy and $E_f= \gamma\; mc^2$
$$\Delta E= \gamma\; mc^2-mc^2=(\gamma -1)mc^2$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}$
$$\Delta E=\left (\left[1-\frac{v^2}{c^2}\right]^{-1/2}-1\right)mc^2$$
Plug the known;
$$\Delta E=\left (\left[1-\frac{0.99973^2c^2}{c^2}\right]^{-1/2}-1\right)(20,000)(3\times 0^8)^2$$
$$\Delta E=\color{red}{\bf 7.57\times 10^{22}}\;\rm J$$
$$\color{blue}{\bf [c]}$$
$$\dfrac{\Delta E}{\Delta E_{\rm Us}}=\dfrac{ 7.57\times 10^{22}}{ 1\times 10^{20}}=\bf 757$$
So the energy needed for such a trip is about 760 times the energy used in the US in 2000, which is an enormous amount of energy.
