jmc

We begin by converting all the numbers to base 10: $$\begin{array}{rl}165_7& =1(7^2)+6(7^1)+5(7^0)=49+42+5=96\\ 11_2& =1(2^1)+1(2^0)=2+1=3\\ 121_6& =1(6^2)+2(6^1)+1(6^0)=36+12+1=49\\ 21_3& =2(3^1)+1(3^0)=6+1=7\end{array}$$Thus, the original expression becomes $\frac{96}{3}+\frac{49}{7}=32+7=\boxed{39}$.