The correct answer is $\boxed{\frac{{\text{g}}}{3}}$.
The centripetal acceleration of the water at the edge of the tank is $\frac{v^2}{r}$, where $v$ is the tangential velocity of the water and $r$ is the radius of the tank. The tangential velocity of the water is equal to the product of the angular velocity of the tank, $\omega$, and the radius of the tank. The angular velocity of the tank is equal to the acceleration of the tank, $a$, divided by the radius of the tank. Therefore, the centripetal acceleration of the water is equal to $\frac{a^2r}{r^2} = a$.
The water will spill out of the tank when the centripetal acceleration of the water is greater than the acceleration due to gravity, $g$. Therefore, the acceleration of the tank must be greater than or equal to $\frac{g}{r}$.
The radius of the tank is $2\text{ m}$, so the acceleration of the tank must be greater than or equal to $\frac{g}{2\text{ m}} = \frac{{\text{g}}}{2}$.
The options are:
- $\boxed{\frac{{\text{g}}}{3}}$: This is the acceleration of the tank when half of the water spills out.
- $\frac{{\text{g}}}{2}$: This is the acceleration of the tank when all of the water spills out.
- $\frac{{2{\text{g}}}}{3}$: This is the acceleration of the tank when the water at the edge of the tank is at the point of spilling out.
- ${\text{g}}$: This is the acceleration due to gravity.
Therefore, the correct answer is $\boxed{\frac{{\text{g}}}{3}}$.