Solution:

There are $3^3 = 27$ equally likely placements. Of these, $3$ pile all coins into one jar (max $3$), and $6$ put one coin in each jar (max $1$). The other $18$ split $2$-$1$ (max $2$). So the expected maximum is $\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{51}{27} = \frac{17}{9}.$ Therefore, the answer is D.