Answer
$1.8 \times10^{22 }\;\rm J$
Work Step by Step
We know that the work needed is given by
$$W=\Delta K=\left(\gamma-1\right)mc^2$$
And hence,
$$W= \left(\sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}}-1\right)mc^2\tag 1$$
Now we need to find the speed of the spaceship $v$ that will give this work.
According to the onboard astronauts, the distance traveled is given by
$$L= \dfrac{ L_{1}}{\gamma}$$
where $L_1$ is the distance traveled according to an observer on Earth.
And w eknow that the time the journey would take relative to the onboard astronauts is given by
$$t=\gamma t_1=1\;\rm year$$
where $t_1$ is the time interval the journey would take according to an observer on Earth.
Therefore, the velocity of the spaceship is given by
$$v=\dfrac{L}{t}$$
Plugging from the previous two formulas;
$$v=\dfrac{ \overbrace{L_{1}}^{6.6c}}{\gamma t}=\dfrac{ 6.6c}{ \sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}} \;t}$$
$$v= \dfrac{6.6c\;\times \;\sqrt{ 1-\frac{v^2}{c^2 }}}{ t}$$
$$v^2= \dfrac{6.6^2c^2\;\times \; \left[1-\frac{v^2}{c^2 }\right] }{ t^2}$$
$$v^2= \dfrac{6.6^2c^2\;\times \; \left[ \frac{c^2\;-\;v^2}{c^2 }\right] }{ t^2}$$
$$44.56 v^2= \dfrac{6.6^2 c^2 }{ t^2}$$
Thus,
$$\dfrac{v^2}{c^2}= \dfrac{6.6^2 }{ 44.56 t^2}$$
Plugging into (1);
$$W= \left(\sqrt{\dfrac{1}{1-\dfrac{6.6^2 }{ 44.56 t^2}}}-1\right)mc^2 $$
Plugging the known;
$$W= \left(\sqrt{\dfrac{1}{1-\dfrac{6.6^2 }{ 44.56 \times 1^2}}}-1\right)\times 36000\times(3\times10^8)^2 $$
$$W=\color{red}{\bf1.8 \times10^{22}}\;\rm J$$