First, we need to change the mixed number to an improper fraction:
[tex]2\frac{5}{6}=\frac{(6\cdot2)+5}{6}=\frac{17}{6}\approx2.83[/tex]Now let's evaluate each of the options:
A.
[tex]\frac{1}{8}\times2\frac{5}{6}=\frac{1}{8}\times\frac{17}{6}=\frac{1\cdot17}{8\cdot6}=\frac{17}{48}\approx0.354[/tex]B.
[tex]2\frac{5}{6}\times2\frac{5}{6}=\frac{17}{6}\times\frac{17}{6}=\frac{17\cdot17}{6\cdot6}=\frac{289}{36}\approx8.02[/tex]C.
[tex]2\frac{5}{6}\times1\frac{5}{8}=\frac{17}{6}\times\frac{13}{8}=\frac{17\cdot13}{6\cdot8}=\frac{221}{48}\approx4.60[/tex]D.
[tex]\frac{5}{6}\times2\frac{5}{6}=\frac{5}{6}\times\frac{17}{6}=\frac{5\cdot17}{6\cdot6}=\frac{85}{36}\approx2.36[/tex]E.
[tex]\frac{6}{5}\times2\frac{5}{6}=\frac{6}{5}\times\frac{17}{6}=\frac{6\cdot17}{5\cdot6}=\frac{17}{5}\approx3.4[/tex]Now, we can conclude that options B, C, and E are greater than 2 5/6.