Low Pass Filters and Bode Plot
An octave is twice (or half) the frequency away from a given point. Eg an octave higher than 100Hz is 200Hz.

You do not need to know the exact value of \$V_{out}\$ or \$V_{in}\$, just the ratio and the transfer function can determine this.


Derive the transfer function.
This is a simple 1st order filter \$\frac{V_{out}}{V_{in}} =  \frac{\omega_o}{s + \omega_o}  \$
Determine the "break frequency" a.k.a. cutoff frequency.
This is straight forward for simple RC low pass filters &amp; you have posted this \$ f_{cuttoff} = \frac{1}{2 \pi R C}\$.
determine the frequency of interest.
This is "two octave's above the cuttoff". Remembering 1 octave is x2.  
Determine the output ratio at this frequency.
Using the transfer function from #1. and replacing s with \$j\omega \$ &amp; then determining the magnitude