Answer: $\mathbf C$
This is nothing but a Harmonic Series.
$$\mathrm{H_n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots+\frac{1}{n} = \sum_{k=1}^n$$
Now, we need to find $$\mathrm {H_{3001}-H_{1000}}$$
For finding out approximate answer:
$\frac{1}{n}$ is monotonically decreasing function, so by using Riemann sum definite integral
$$\int_{1002}^{3002}\frac{1}{t}dt \leq \sum_{1001}^{3001}\frac{1}{n}\leq \int_{1001}^{3001}\frac{1}{t}dt$$
which simplifies to:
$$\ln \frac{3002}{1002} \leq \;\text{answer}\; \leq\ln\frac{3001}{1001}$$
This values can be approximated to:
$$1 \lt\;\text{answer}\;1.5 \implies 1 \lt X \lt\frac{3}{2}$$
$\therefore \mathbf C$ is the right answer.