The ammeter resistance is zero, and the battery is ideal

let current through resistance R is I.

from above figure, the voltage across ammeter zero

thus the current in the bottom resistors are the same.

hence, the through the battery is 2I.

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the voltage drop across each of bottom resistors is

                     V_R = IR     (from Ohm's law, V = IR)

the equivalent resistance of the circuit is

                   R_eq  = {[(2.96R)(R)] / [(2.96)R+R]}+{(R)(R)/(R + R)}

                          = [1.2474]R      ...... (1)

the net current throuh the battery is

                      2I = ξ /R_eq

substitute the equation (1) in above equation, we get

                      2I = ξ /(1.2474)R

                        I = (0.4008)[ξ /R]      ...... (2)

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using the loop rule for L.H.S loop,

               ξ -I_2R(2.96R) - IR = 0

                    I_2R = [ξ - IR] /(2.96R)

substitute the equation (2) in above equation, we get

                    I_2R = (0.2024)[ξ /R]      ...... (3)

therefore, the current through the ammeter is

                    I_A = I - I_2R   

                        = [(0.4008)ξ /R] - [(0.2024)ξ /R]   

                        = (0.1983)[ξ /R] 

       I_A / [ξ /R]  = 0.1983