Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that
$$E_n=\dfrac{n^2h^2}{8mL^2}$$
The droplet is moving as a particle in a one-dimensional rigid box. Sp, its energy is entirely kinetic.
Hence,
$$K=\dfrac{n^2h^2}{8mL^2}$$
Solving for $n$ to find the quantum number.
$$n=\sqrt{\dfrac{8mL^2 K}{h^2}}=\dfrac{L\sqrt{8m K}}{h}$$
Recalling that $K=\frac{1}{2}mv^2$, so
$$n =\dfrac{L\sqrt{8m (\frac{1}{2}mv^2)}}{h}=\dfrac{2mv\;L}{h}$$
Now we need to find the mass of the droplet which is given by $m=\rho V=\frac{4}{3}\pi r^3 \rho$, so
$$n =\dfrac{2(\frac{4}{3}\pi r^3)\rho v\; L}{h}$$
$$n =\dfrac{8\pi r^3 \rho\; v\;L}{3h}$$
Plug the known;
$$n =\dfrac{8\pi (1\times 10^{-6})^3 (1000)\; (1\times 10^{-6})\;{(20\times 10^{-6})}}{3(6.63\times 10^{-34})}$$
$$n=\color{red}{\bf 2.53 \times 10^8}$$
$$\color{blue}{\bf [b]}$$
The correspondence principle states that for large quantum numbers, the behavior of quantum systems should approach classical behavior. We can use this principle to determine whether quantum mechanics is required or if classical mechanics is sufficient to describe the particle's motion.
In other words,
If the quantum number $n$ is very large, we need to use classical mechanics.
If the quantum number is small, we need to consider quantum mechanics to understand the motion of the particle.
Generally, if $n \gg 1$, the particle's behavior is classical.
The calculated quantum number $ n $ is extremely large.
($n\approx\bf 2.53\times 10^8 $), which means the particle behaves classically. According to the correspondence principle, since $ n \gg 1 $, the particle's motion can safely be described using classical physics, and quantum mechanics is not necessary in this case.
