I'm going to assume you were just handed this schematic and asked to perform some tasks. The results you got do not sound correct for the circuit. (It's not a great design. But it's enough to get by.)
The first thing to check is the voltage of the power supply. If this is a \$9\:\text{V}\$ battery that was sitting on a lab shelf, definitely check it out with a meter -- using a \$1\:\text{k}\Omega\$ resistor across its terminals while you use the meter to measure the voltage. If this is a commercial power supply, check it anyway. It doesn't hurt to verify that your voltage is what it is supposed to be.
The next thing to check is the resistor values. Make sure they are near the values you are told they should be. Again, a few moments spent here can save you lots of time, later. It's only five resistors, so it shouldn't take a lot of time.
It's a good idea to have checked the datasheet for the specific BJT you are using. The 2N3904 comes in a variety of packages and I wouldn't be surprised to find different pinouts even for the same TO-92 package. It's worth a moment to make sure you know where your pins are located. I won't suggest that you spend time checking out the BJT, since you will be doing that soon enough when you plug it into the circuit. But if you do have a transistor checker available, please do use it.
Finally, check your wiring two or three times over. I make mistakes. So you can make them, too. Check things over several times. Might save more time, later.
I won't explain, but I'll do a few quick calculations for you. These will be added here so that you can check this out with a voltmeter before running your tests.
Assuming that the BJT isn't saturated, the base current should be:
$$I_\text{B}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\:R_\text{E}}$$
Here, \$R_\text{E}=200\:\Omega+2.2\:\text{k}\Omega=2.4\:\text{k}\Omega\$, \$R_\text{TH}=\frac{R_1\:R_2}{R_1+R_2}\$ and \$V_\text{TH}=9\:\text{V}\frac{R_2}{R_1+R_2}\$. The value for \$V_\text{BE}\$ will vary a little, but I can already tell that it's closer to \$600\:\text{mV}\$ than to \$700\:\text{mV}\$. So I'll just pick something just under the middle of that. This gives me about \$800\:\text{nA}\$ for the base current (using a random guess for \$\beta\approx 260\$.) And therefore a collector current of only about \$210\:\mu\text{A}\$. About a \$2.1\:\text{V}\$ drop across the collector resistor.
So the next thing you do is do NOT put a signal up to your circuit. Just leave it open for now. Apply power and measure the output voltage. It should be in the area of about \$6.5\:\text{V}\$ to \$7.3\:\text{V}\$. Hopefully nearer the middle of that area. If the voltage is outside that range, stop everything and go measure the base voltage. This should be:
$$V_\text{B}=V_\text{TH} - I_\text{B}\cdot R_\text{TH}$$
In this case, that works out to about \$1.1\:\text{V}\$ to \$1.2\:\text{V}\$. If the base voltage also is too far outside that range (a few tenths of a volt above or below it), then just stop. Something is wired wrong. Get help and/or check things out.
These are basic checks.
If these things seem okay, then there is a good chance you have it wired correctly. Go back and verify the voltage source, just in case. Make sure it's still \$9\:\text{V}\$.
If all that is in line, then go ahead and add the capacitors and then apply your starting signal. You definitely should be getting more than a few millivolts (at the collector, relative to the [-] terminal of the power supply.)
Which reminds me -- polarity is important. Make sure that the emitter of the BJT is towards the [-] side of the power supply and that the collector of the BJT is towards the [+] side.
My guess is that you didn't hook up the power supply correctly, got a PNP instead of an NPN, or really flubbed up the wiring somehow. Check things. Then check them again.
Also, this particular circuit will have varying gain -- and some distortion as a result. (If you actually get it built up correctly.) But for these purposes, that should be fine enough.
