Dear...

Question is very simple

..

See specification of ammerter is 9 ohm and 10 mA

That means it can have maximum current 10 mA with it

So from I which enters through A ...10 mA will goes through Ammeter and then 0.9 ohm resistance and finallt leave through terminal B

Also the net current (I-10) will pass through resistance 0.1 ohm and leave from B

SO final current entering A is I and leaving b is (I-10)+10

So The potential accross AB will be equal to potential accros terminal of Ammeter and accross resistance 0.9 ohm

So aply equation

9(10mA) + 0.9(10mA) = (I-10)mA ( 0.1)

99 = (0.1* I - 1 )

so 0.1*I = 100

or I = 1000 mA

or I = 1 A

Note : for furthure doutb ask !

Good wishes