I'll start by telling you the complete way to calculate the noise in the measurement, and then I'll tell you how to make a quicker estimate.
Lock-in Amplifiers: First of all, it is helpful to explain how a lock-in amplifier works. It works by taking the input signal and demodulating it against the internal (or externally supplied) sine wave reference. If your input signal is $V_i = A_i\cos(\omega_i t+\phi)$ and the internal reference is $V_r=\cos(\omega_r t)$, then the demodulated signal is given by $$ \begin{align} V_d &= \cos(\omega_r t)\ A_i\cos(\omega_i t + \phi) \ &= \frac{A_i}{2}\left[\cos((\omega_i-\omega_r)t - \phi) + \cos((\omega_i + \omega_r)t + \phi) \right] \end{align} $$ Usually $\omega_i\approx\omega_r$ because the lock-in's frequency reference is also used to drive the system being measured. In this case the first term is near DC while the second term has twice the frequency of the driving signal. After the demodulation, the lock-in amplifier low-pass filters the signal to get rid of the second term leaving only $$ V_d=\frac{A_i}{2}\cos((\omega_i-\omega_r)t - \phi). $$ The bandwidth of this low pass filter is what sets the frequencies you are able to see as well as the noise in the measurement.
Full Noise Calculation: The RMS noise is usually calculated from the PSD (PSD = ASD$^2$) with an extension of <PERSON>'s theorem which says that the RMS variance of a time domain signal is given by $$ \sigma^2=\int_0^\infty PSD(f)df $$ In the case of a lock-in amplifier you have to take the low-pass filter into account by multiplying by the shifted, double sided filter. 552
Once you have the variance, the standard deviation is just the square root of the variance, and the noise in your measurement after averaging over many cycles is given by the standard error of the mean $$ SE = \frac{\sigma}{\sqrt{N}} $$ where $N$ is the number of cycles that you average.
Quick Estimate: Now for what you are probably most interested in. You can make a quick estimate of the noise at a given frequency by simply taking the value of your measured noise at that frequency and multiplying by the square root of twice the bandwidth: $$ \sigma(f)\simeq ASD(F)\times\sqrt{2\ BW}. $$ The noise is you measurement after averaging over many cycles is then given by $$ SE(f)=\frac{\sigma(f)}{\sqrt{N}}. $$ So, for your example of 10 Hz, assuming that the bandwidth of your lock-in filter is set to 1 Hz, the noise would be approximately $$ \sigma(10\ Hz) = 600 \frac{nV}{\sqrt{Hz}}\ \sqrt{2\ Hz}=850\ nV. $$ If you average for 100 cycles, then the noise comes down to $$ SE(10\ Hz) = \frac{850\ nV}{\sqrt{100}}=85 nV. $$ 552