1 kHz square wave into a 6.8k resistor in series with a 10 nF capacitor to common ground.
I was doing this very thing last week as part of a calculus demonstration of v(t)=v0(1-e-t/T) where v0 is the applied voltage. The calculus part of the deal was to find the inverse function: t in terms of v(t)/v0, a fraction less than or equal to 1.
T is tau, the RC time constant and you should show 6 time constants. So, 6.8*103 * 10*10-9 = 68 uS. Then 68 uS * 6 (tau) -> 408 uS. Figuring both sides of the square wave we need 816 uS or just a little less than 1 mS or about 1 kHz.
Still have the breadboard sitting on my bench...