Number of terms in the expansion is $10+1=11$
Middle term of the expansion is $\large\frac{11+1}{2}$$=T_6$
$T_{r+1}=C(10,r)(\large\frac{x}{3})^{10-r}$$(9y)^r$
But $T_{n+r}=T_6$ or $r+1=6$
Putting $r=5$ in (1) we have
$T_6=T_{5+1}=C(10,5)(\large\frac{x}{3})^{10-5}$$(9y)^5$
$\Rightarrow C(10,5)\large\frac{x^5}{3^5}$$.9^5y^5$
$\Rightarrow C(10,5) 3^5 X^5y^5$
$\Rightarrow \large\frac{10!}{5!(10-5)!}$$3^5x^5y^5$
$\Rightarrow \large\frac{10!}{5!5!}$$3^5x^5y^5$
$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}$$3^5x^5y^5$
$\Rightarrow 61236 x^5y^5$
Hence (C) is the correct answer.