For two weightings it's quite simple:
- Take 1 ball from every machine,
ActualWeight - 10gr*10 is the difference for the faulty machine
- Using information from #1 you have the original task which is solvable with 1 weighting
I believe 1 weighting is not sufficient and here is why: if you take n1,n2,n3,...,n10 balls from corresponding machines, it's equally possible that 1st machine is faulty with weight difference 1/n1 and that 2nd machine is faulty with weight difference 1/n2 - these cases don't seem to be distinguishable...
If you ask for $13,13,13$, then you get nothing if the boxes are $10,8,12$.
If you ask for $12,12,12$, you will always get exactly $12$ coins, since in all cases, exactly one of the boxes has at least $12$ coins.
Claim $12$ is best possible.
Certainly it's best possible with all requests equal.
Suppose a better request triple is $a,b,c$, with $a \le b \le $c, and $a < c$.
But any of the $3$ requests might fail if that box contains $0$ coins, hence to be sure to get more than $12$ coins, every pair must sum to more than $12$.
From $a \le b \le c $, and $a + b > 12$, we get $6 < b \le c$.
But if the boxes are some permutation of $26,0,4$, the $b,c$ requests might both fail, so to be sure to get more than $12$, we must have $a > 12$.
But then $a,b,c$ all exceed $12$, so all requests would fail if the boxes are $10,8,12$.
Thus, unequal requests can't beat the uniform $12,12,12$ request strategy.
It follows that $12$ is the maximum number of coins which can be guaranteed.
Best Answer
Now with the edit, this is solvable. Choose the BW tagged box. All boxes are wrongly tagged by definition. Pick a ball, if ball is black, then, this box is not WW for sure, hence it is BB. Now if the BB-tagged box were originally BW box, then the WW-tagged box would be rightly tagged, which would be a contradiction, hence, BB-tagged box is actually a WW box, and WW tagged box is actually a BW box.
Similarly, you can retag boxes if a white ball came out of BW box.