The force on the body is given as \( \vec{F} = 6 \hat{k} \, \text{N} \), which is directed along the positive z-axis.
Using Newton's second law \( \vec{F} = m \vec{a} \), where \( m \) is the mass of the body:
\[
\vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \, \text{ms}^{-2}
\]
Thus, the acceleration of the body is \( 3 \hat{k} \, \text{ms}^{-2} \), which means the body accelerates in the positive z-direction.
To find the final velocity, we use the equation of motion:
\[
\vec{v}_{\text{final}} = \vec{v}_{\text{initial}} + \vec{a} \Delta t
\]
Substituting the given values:
\[
\vec{v}_{\text{initial}} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1}, \quad \vec{a} = 3 \hat{k} \, \text{ms}^{-2}, \quad \Delta t = \frac{5}{3} \, \text{seconds}
\]
\[
\vec{v}_{\text{final}} = (3 \hat{i} + 4 \hat{j}) + (3 \hat{k}) \times \frac{5}{3}
\]
\[
\vec{v}_{\text{final}} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}
\]
Thus, the final velocity of the body when it emerges from the force field is:
\[
\boxed{3 \hat{i} + 4 \hat{j} + 5 \hat{k}}
\]