Electronic – Multimeter, measuring current calculates less. Can I calculate meter’s internal resistance this way

I have set up the following circuit in order to measure current.

^{simulate this circuit – Schematic created using CircuitLab}

Of course, Ohm's law tells me that I should get a measurement of about 0.3A (300mA).

3V / 10 Ohm = 0.3

However, my meter shows me a value that is much smaller? Actually, it's about half that at 157.7mA

Other Things I've Considered

Since I've tested my battery voltage and it is 3 to 3.01 volts I'm assuming it is fine. I also tested the 10Ohm in resistor with the meter and the meter measures it at 9.8 to 10 Ohms which seems correct.

Meter's Internal Resistance

After a couple of searches I believe this is due to the meter's internal resistance.
If that is true then I'm calculating the internal resistance at about 10 Ohms also since:

3V / 0.157(mA) = 19.10 (Ohms)

Raises A Few Questions

But I still have a few questions.

1. From now on when I measure current with this meter should I just consider that I have about 10 Ohms of extra resistance? Will this value be consistent? Is that a valid assumption?
2. If this is a known value why didn't the meter maker/designer just subtract that value from the calculation they do every time so the meter would display the correct value?
3. Am I missing some other piece of vital information (i.e. this is not the meter's internal resistance)?
4. Is it possible to buy a meter which would measure the current accurately* — closer to 300mA? Would it be cost prohibitive for a hobbyist?
5. If you can buy a meter like that, what might the description show that would help me to know it could measure current without being affected by internal resistance?

*I know this is somewhat ridiculous because once you know, you know. However, I'm attempting to teach Ohm's law to early electronics students and attempting to show them that the meter will confirm their calculations.

EDIT
Because there were some questions about my initial voltage measurement I did some more investigation.

My original voltage measure was on the two AA batteries with no load.
I then measured approx 2.2 volts across the 10 Ohm resistor in my circuit.
I'm assuming that would mean something like:

2.2 / 10 Ohm = 220mA

However, I know when I measure that I only see 157mA of current.
So…

220mA - 157mA = 63mA (which are missing somewhere).

I plug in the missing 63 in the following way:

.8V (voltage drop) / .063 = 12.69 Ohms

So I'm still seeing a extra ~12 Ohms that I can't account for. That is obviously higher than my 9-10 Ohm extra I was calculating but could be due to rounding.

Does this add some additional valuable information? I think so.

EDIT 2

Here's a zoomed in snapshot of the meter panel because a helpful post(er) thought the meter was in AC mode. As you can see, the meter was in DC mode during the measurement.