Simple Circuit Giving Weird V=IR Calculation
Another way to look at it would be like this:

$$ V/I = R $$

V is known and so is I \$ 1.61V/0.22A = 7.3Ω \$ which is probably the resistance of your battery, meter and wires in between. You are probably shorting out the resistor by measuring across the resistor in current mode. 

While measuring in current mode you need to place the meter in series with the resistor, not measure across it. Since I'm not at your bench I don't know exactly what is going on. 

Another way to make the current measurement is use the resistor and measure the voltage across it. 

$$ V_{res}/R = I  $$

which is not going to be exactly the same as the voltage across your battery (there are small losses from the cables) but it will be close.

It should be something like \$ 1.61V/10300Ω = 0.000156A\$ or 156uA

Keep in mind that your batter voltage will drop when more current is drawn from it (it has internal source resistance, batteries are not ideal voltage sources).