Quantity left after $n$ operations $=x\left(1 -\dfrac{y}{x}\right)^{n}$
where, $x=$ initial quantity
$y=$ amount of mixture withdrawn each time (this should be same every time)
$n=$ no. of times operation performed
$\quad = 10\left(1 -\dfrac{1}{10}\right)^{n}$
$\quad =10 \left(\dfrac{9}{10}\right)^{3}$
$\quad = 10\times 0.9\times 0.9\times 0.9$
$\quad = 10\times 0.729 = 7.29$ litres
Hence, option $D$ is correct.