Answer
$0.3nm$
Work Step by Step
The resolving power $R$ of a diffraction grating is given by (in the book):
$$R=Nm$$
where $N$ is the number of slits in the grating and $m$ is the order of diffraction.
Furthermore, the resolving power $R$ is related to the wavelength incident on the grating by
$$R=\frac{\lambda_{avg}}{\Delta\lambda}$$ where $\lambda$ is the average of the wavelengths of the two emission lines and $\Delta\lambda$ is the minimum difference of wavelength of the two emission lines.
Hence it can be written as
$$R=\frac{\lambda+\lambda'}{2(\lambda-\lambda')}$$
$\lambda$ is the incident wavelength and $\lambda'$ is the wavelength of the second emission line. The numerator is the explicit mean of the two emission line wavelengths.
In this question, $N=1000*2=2000$ gratings and $m=2$. Hence,
$$R=2000=\frac{600nm+\lambda'}{2(600nm-\lambda')}$$
Hence, $$4000(600-\lambda')=600+\lambda'$$
Solving this gives
$$2400000-4000\lambda'=600+\lambda'$$
Therefore,
$$\lambda'=599.7 nm$$
Thus the smallest wavelength difference that can be resolved in the second order by this grating is $$600-599.7=0.3nm$$
