$\(r=2.666666 \ldots\)$ as a repeating decimal equals $\(\frac{8}{3}\)$, so $\(a=8\)$ and $\(b=3\)$, giving $\(a+b=11\)$, which is greater than 10 , so Quantity $A$ is greater.

Converting to a fraction
Let $\(r=2 . \overline{6}\)$, meaning the digit 6 repeats forever.

Then $\(r=2+0 . \overline{6}=2+\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}\)$.

Thus $\(a=8, b=3\)$, and $\(a+b=8+3=11\)$, which is greater than 10 , so Quantity $\(\mathrm{A}>\)$ Quantity B .