Value of C in the RC Circuit
The RC circuit in the question is not an integrator but, it can behave like one, providing the output level is low enough (compared to the input level) and, this only happens, for continuous signals, when the frequency applied is significantly greater than the low-pass cutoff frequency. Consider the transfer function of the circuit and compare it with an integrator: -



The 3dB point is when the amplitude response is at 0.7071 and this happens when \$\omega_0 = \dfrac{1}{RC}\$. 

And now the integrator: -



The difference is that the gain of the integrator at low frequencies continues to rise as frequency drops. That cannot happen with a low-pass RC network and so only the sloping part of the line mimics an integrator as it tends to become close to attenuating at 20 dB per decade.

You could make a case that this happens at twice \$\omega_0\$ or you could be more picky and say it doesn't really get accurately close until 10x \$\omega_0\$. You could be a perfectionist and say 100x or 1000x.

So, what margin of error is acceptable? Do I know? No of course I don't know - there is no theoretical point at which it can be regarded as behaving exactly like an integrator - it just gets closer and closer as frequency rises.

So what is the answer the OP should give in his homework? He should, I believe make a case for 2x, 10x and 100x the frequency and explain what the error is if a sinewave was the signal inputted. Alternatively you could make a case for a square wave being inputted and look how linear the resulting triangle wave will be. It's his call.

BTW it's the same for an RC differentiator - it's not a perfect differentiator.