Now when I measure current across the first and the second resistor individually I get around 20mA
That is correct, because (as highlighted in a comment by jsotola) each time you try to measure current "across" an individual resistor, you are shorting it out. Therefore for each of those two measurements, the only resistor still in the circuit is the other 100 Ω resistor and so the nominal current is 1.87 / 100 = 18.7 mA (assuming those exact values for V and R). That is why you are measuring approximately 20 mA.
See my answer here explaining a very similar mistake by someone else.
Don't try to measure the current "across" both resistors, as then you are shorting out both of them, so neither of them is limiting the current from the battery, and the only limit becomes things like the internal resistance of the battery and the wiring. It is easy to damage your multimeter, or blow its internal fuse on that range (if it has one).
If you are using a multimeter on its current range, it must be in series with the circuit whose current you are trying to measure (you are using it in parallel - that is the mistake). Again, see the link above.