Given m( ABC) = 60^ and m( ACB) = 40^:

The sum of angles in a triangle is $180^\circ$. In $\triangle ABC$:

$$m(\angle BAC) + m(\angle ABC) + m(\angle ACB) = 180^\circ$$

Given $m(\angle ABC) = 60^\circ$ and $m(\angle ACB) = 40^\circ$:

$$m(\angle BAC) + 60^\circ + 40^\circ = 180^\circ$$ $$m(\angle BAC) + 100^\circ = 180^\circ$$ $$m(\angle BAC) = 180^\circ - 100^\circ$$ $$m(\angle BAC) = 80^\circ$$

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle. In this case, $\angle BOC$ is the central angle subtended by arc BC, and $\angle BAC$ is the inscribed angle subtended by the same arc BC.
Therefore:

$$m(\angle BOC) = 2 \times m(\angle BAC)$$ $$m(\angle BOC) = 2 \times 80^\circ$$ $$m(\angle BOC) = 160^\circ$$

The final answer is \boxed{\text{160^\circ}}.