Given that, $9^{2x-1} – 81^{x-1} = 1944 $
$\Rightarrow 9^{2x-1} – 9^{2x-2} = 1944 $
$\Rightarrow 9^{2x}\cdot 9^{-1} – 9^{2x}\cdot9^{-2} = 1944 $
$\Rightarrow 9^{2x}\left(\frac{1}{9}-\frac{1}{81}\right) = 1944 $
$\Rightarrow 9^{2x}\left(\frac{8}{81}\right) = 1944 $
$\Rightarrow 9^{2x} = \frac{1944}{8} \cdot 81$
$\Rightarrow (3^{2})^{2x} = 243 \cdot 81$
$\Rightarrow 3^{4x} = 3^{5} \cdot 3^{4}$
$\Rightarrow 3^{4x} = 3^{9}$
The base is same on both sides, then equating the powers.
$\Rightarrow 4x = 9 $
$\Rightarrow \boxed{ x = \frac{9}{4}}. $
Correct Answer $:\text{B}$