How to find the power absorbed by independent voltage source if there is an independent current source
Having measured the current through R1, you now have enough information to use KCL to find the current through R2. And the current through V1 is the same as the current through R2.
Knowing the current through R1 you also have enough information to find the voltage at the unlabeled node connecting R1, I2, and R2, and thus to find out the voltage of R1.
No you just use \$P=IV\$ to find the power absorbed by V1.

the power absorbed by the voltage source (which I think should be negative, considering voltage sources deliver power not absorb them)

The power absorbed by the voltage source can be either positive or negative.
For example, if you are charging a battery, it will absorb power, not deliver it.
Or, if you model a diode with forward bias as a voltage source, it will always absorb power.
Or if you simply design a random circuit with ideal components where the current happens to flow in to the positive terminal of the voltage source, then it will absorb power.
Example
To make it concrete, I'll work an example. But I'll choose a different current at the meter in case the values you gave were the ones assigned to you to work out as homework.
Let's say the meter reads 3 A.
Then we know there must be 2 A flowing left to right through R2 (because of KCL).
And we know the voltage at the top of R1 must be 90 V (because of Ohm's law).
Then the voltage at the top of the voltage source must be
$$V_1 = 90\ {\rm V} - (2\ {\rm A})(10\ {\rm\Omega})= 70\ {\rm V}$$
And the power absorbed by the voltage source is
$$P = (2\ {\rm A})(70\ {\rm V}) = 140\ {\rm W}$$
Notice this is positive, because the current source is producing a higher voltage than the voltage source is, and so it is sending current back toward the voltage source.