$\left(\frac{1}{60}-\frac{x^{8}}{81}\right)\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$

Term independent of $x$ will be $\frac{1}{60} \times$

independent of $x$ in $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{1}{8} \times$

Termof $x^{-8}$ in $\left(2 x^{2} \frac{3}{x^{3}}\right)^{6}$

$\mathrm{T}_{r+1}$ in $\left(2 \mathrm{x}^{2}-\frac{3}{\mathrm{x}^{2}}\right)^{6}$ will be

$\mathrm{T}_{r+1}=^{6} \mathrm{C}_{r}\left(2 \mathrm{x}^{2}\right)^{6-\mathrm{f}}\left(-\frac{3}{\mathrm{x}^{2}}\right)^{r}$

$=^{6} \mathrm{C}_{r} 2^{6-5}(-1)^{r} \times 3^{r} \times \mathrm{x}^{12-2 r-2 r}$

Case $I:$ For term independent of $x, 12-4 r=0$ $\Rightarrow r=3$

$T_{4}=^{6} C_{3} \times 2^{3} \times 3^{3} x^{6}=-20 \times 2^{3} \times 3^{3}$

case $II :$ For term of $x^{-8}$

${12-4 r=-8}$

${4 r=20 \Rightarrow r=5}$

$T_{6}=^{6} C_{5} \cdot 2^{1}(-1) \cdot 3^{5} \cdot x^{-8}$

Required Answer $=\frac{1}{60} \times(-20) 2^{3} \times 3^{3}-\frac{1}{81} \times 6 \times 2 \times(-1) \times 3^{5}$

$=-72+36=-36$

Hence the correct answer is option $(B).$