Ans: A
Since $h(x)$ is divisible by $2x-3$, we have
$\begin{array}{rcl}
h\left(\dfrac{3}{2}\right) & = & 0 \\
a\left(\dfrac{3}{2} \right)^6+16\left(\dfrac{3}{2}\right)^3+b & = & 0 \\
\dfrac{729a}{64} +b+54 & = & 0 \\
\dfrac{729a}{64}+b & = & -54 \ldots \unicode{x2460}
\end{array}$
The remainder when $h(x)$ is divided by $2x+3$
$\begin{array}{cl}
= & h\left(\dfrac{-3}{2}\right) \\
= & a\left(\dfrac{-3}{2}\right)^6 +16\left(\dfrac{-3}{2}\right)^3 +b \\
= & \dfrac{729a}{64} +b -54 \\
= & -54-54 \text{ , by $\unicode{x2460}$} \\
= & -108
\end{array}$
