Let the rowing rate of the boat crew be x. When they row upstream they row against the rate of flow of the stream, but when they row downstream, their rowing rate is improved by the stream flow rate. Upstream their actual speed is x-3 and downstream it's x+3. The time to row upstream is t1=6/(x-3), and the time to row downstream is t2=6/(x+3). t1+t2=8/3 hrs, so 6/(x-3)+6/(x+3)=8/3. Divide through by 6: 1/(x-3)+1/(x+3)=4/9.
(x+3+x-3)/(x^2-9)=4/9; 2x/(x^2-9)=4/9; x/(x^2-9)=2/9; multiply by 9(x^2-9): 9x=2(x^2-9)=2x^2-18.
We now have a quadratic equation: 2x^2-9x-18=0, factorising to: (2x+3)(x-6)=0. So x=6 km/hr. (The other solution gives -3/2 km/hr, which means they would be rowing in the opposite direction and would give a negative time for t1, which must be rejected.)
Check: t1=6/3=2 hrs; t2=6/9=2/3 hrs, so total time is 2 2/3 hrs. OK!