Call the pipes X and Y and let the amount of time for X to fill the tank be x and the time for Y to fill the tank x-3. In 1 hour X fills 1/x of the tank and Y 1/(x-3) of the tank. Together they fill 1/x+1/(x-3) of the tank. 6hr 40min=6 2/3 hr=20/3hr. In one hour both pipes fill 3/20 of a tank, so 1/x+1/(x-3)=3/20. Multiply through by 20x(x-3): 20(x-3)+20x=3x(x-3); 20x-60+20x=3x^2-9x; 3x^2-49x+60=0; (x-15)(3x-4)=0; x=15 or 4/3 and x-3=12 or -5/3, so we accept only the positive solution: the two pipes fill the tank in 15 and 12 hours respectively.
Check: 1/15+1/12=4/60+5/60=9/60=3/20.