$\big(\large\frac{1}{a}-\frac{2b}{3}\big)^{10}$
$T_{r+1}=(-1)^r 10C_r (\large\frac{1}{a})^{10-r} (\large\frac{2b}{3})^r$
$\Rightarrow (-1)^r 10C_r (\large\frac{1}{a})^{10-r}.\large\frac{(2b)^r}{3^r}$
$T_5=(-1)^r 10C_4(\large\frac{1}{a})^6(\frac{2b}{3})^4$
$\Rightarrow (-1)^5 10C_4 \large\frac{1}{a^6} \frac{2b^4}{3^4}$
$\Rightarrow - 10C_4 a^{-6 }\large\frac{2b^4}{81}$-----(1)
$10C_4=\large\frac{10!}{4! 6!}$
$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$
$\Rightarrow 210$-----(2)
Substituting (2) in (1) we get
$-210\times a^{-6} \times \large\frac{16b^4}{81}=\frac{-1120}{27}$$a^{-6}b^4$
Hence (C) is the correct answer.