Ans: $a=\dfrac{-2}{9}$, the coefficient of $x^2=-248$
$\begin{array}{cl}
& (2-3x)^5\left(x+\dfrac{a}{x}\right)^2 \\
= & [C^5_12^5+C^5_1(2)^4(-3x)+C^5_2(2)^3(-3x)^2+C^5_3(2)^2(-3x)^3+C^5_4(2)(-3x)^4+C^5_5(-3x)^5] \\
& \ \ \ \ \ \left(x^2+2a+\dfrac{a^2}{x^2}\right) \\
= & (32-240x+720x^2-1080x^3+810x^4-243x^5)\left(x^2+2a+\dfrac{a^2}{x^2}\right)
\end{array}$
Consider the coefficient of $x$, we have
$\begin{array}{rcl}
(-240)(2a)+(-1080)(a^2) & = & \dfrac{160}{3} \\
3240a^2+1440a+160 & = & 0 \\
81a^2+36a+4 & = & 0 \\
(9a+2)^2 & = & 0
\end{array}$
$\therefore a=\dfrac{-2}{9}$ (repeated).
The coefficient of $x^2$
$\begin{array}{cl}
= & (32)(1)+(720)[2(\dfrac{-2}{9})]+(810)\left(\dfrac{-2}{9}\right)^2 \\
= & -248
\end{array}$
