5.6 years
=
According to [this page][1], distance travelled under constant force acceleration, even up to relativistic speeds, is calculated by:

$s(t) = c(\frac{m_0c}{F})(\sqrt{1 + (\frac{F}{m_0c})^2t^2} - 1)$

...and since...

 $\frac{m_0}{F} = a = g \approx 10 m/s^2$

...and since we accelerate half-way and decelerate halfway we calculate the time to travel half the distance and then double that. Half the distance is...

$s=300\cdot10^6m/s\cdot60s/min\cdot60min/h\cdot24h/day\cdot365day/y\cdot2y \approx 19\cdot10^{15}m$

...so from this we get...

$19\cdot10^{15}m = 3\cdot10^8\cdot\frac{3\cdot10^8}{10}\cdot(\sqrt{1 + \frac{10^2}{3\cdot10^8\cdot3\cdot10^8}t^2} - 1)$

...which gives us...

$t = 88\cdot10^6s = 2.8y$

Double this and you get a total traveltime of **5.6 years**.


  [1]: https://xphysics.wordpress.com/2010/11/07/relativistic-acceleration-due-to-a-constant-force/