a= 1.3mm \(\pm\) 0.1mm so a is between 1.2 and 1.4mm
b= 2.8mm \(\pm\) 0.2mm b is between 2.6 and 3.0mm b^2 is between 6.76 and 9.0 mm^2
c= 0.8mm \(\pm\) 0.1mm c is between 0.7 and 0.9mm sqrt(c) is between 0.836 rounded down and 0.987 rounded up
Calculate \(Q=\frac{ab^2}{\sqrt{c}}\) with its uncertainty.
ab^2 is between 1.2*6.76 and 1.4*9 mm^3 ... ab^2 is between 8.112 and 12.6 mm^3
ab^2/sqrtc is between 8.112/0.987 and 12.6/0.836 mm^3/mm^0.5.......
ab^2/sqrtc is between 8.218 rounded down and 15.072 rounded up mm^2.5
(8.218+15.072)/2 = 11.645
so the average is 11.64, the smallest is 11.645-3.427=8.218 and the biggest is 11.645+3.427=15.072
So that is 11.645 pm 3.427 approx but this degree of accuracy is not valid and I am not sure what would be valid.
I'd say maybe
11.6 \pm 3.5 OR 12 \pm 4 would be reasonable.
I'm going with 12 pm 4 but I expect there is a more defined way of chosing the degree of uncertainty that is scientifically valid.