The problem description doesn't show the circuit between Vin and the transistor's base, so you can't make any assumptions as to what the logic for turning it OFF is.
The only thing that is clearly mentioned is that the transistor will be turned ON when Vin reaches 3.4V. We can safely assume that the circuit will not turn the transistor back off as soon as the voltage goes a bit under 3.4V, since then the circuit would just oscillate around that point and never discharge the capacitor. We can also conclude, given the wording used, that the transistor will be turned on fully (i.e. into saturation).
There are many possible circuits that would achieve such an effect. Depending on which is used, you can make the circuit turn off the transistor at any point you wish... 1V, 2V, etc., or never turn it off at all.
If you replace the cap with a resistor.. initially Vin will be the result of the voltage divider. If it's <<3.4V obviously nothing will happen and the circuit will stay that way forever. If it's more than that, the transistor will be turned ON and basically pull Vin close to 0 almost instantly. What happens then is again unspecified. As before, depending on the circuit, it might be turned off and restart the cycle at high speed (thus the whole thing becomes an oscillator), or it might just stay open and keep drawing current through the top resistor.