A critical point in your description of the problem — a point which you
neglected to consider — was to NEGLECT WATER RESISTANCE!! In fact, you
have made the water resistance substantial, because you have the man
stopping at x=12! If the man is standing on the end of the barge at x=12,
that means the barge didn’t move at all! If the barge does move — even a
little bit (and it does!), the end of the barge WON’T BE AT X=12! Either the
man stops BEFORE x=12 or he has to keep walking PAST x=12 (depending upon
which way the barge moves) in order to be at the stern when he stops.

So here’s what is happening…

Since water resistance is neglected, the man-barge system is isolated from
external forces. That means its center of mass remains STATIONARY (since
everything is initially at rest). Therefore, when the man walks to the
right, the barge MUST move to the left in such a way that the center of mass
REMAINS at 5.65. (The man must push BACK against the floor of the barge in
order to walk forward, and since we are neglecting water resistance, the
backward push results in backward barge motion that balances (with regard to
center of mass) the man’s forward motion.

You can get your answer just by considering the symmetry of the situation.
When the man stops at the stern of the barge, he will be 5.65m away from the
center of mass, just as he was when he was at the bow. But the center of
mass must still be at 5.65 (because there are no forces external to the
man-barge system). So that means the man is at 11.30m and the front of the
barge must be 12m to the left of that; namely, 11.30 – 12 = -0.7m. So the
barge moves .7m to the left.

Just as a double check, compute the new center of mass:

(70×11.29 + 1120×5.3)/1190 = 5.65