what multiple of ε/r gives the current in the ammeter?

The most salient feature of this problem is to notice that, since the ammeter has no internal resistance, the two lower resistors are in parallel, i.e. the voltages across each of them are the same. This is also true for the two resistors on top. Since the two lower resistors have the same resistance, the fact that the have the same voltage implies that they also carry the same current.

 Name this correct i. Since these two currents join at the bottom and go to the battery on the left, the full current through the battery is 2i. Using this fact, and computing the equivalent resistance which turns out to be Req = 7/6R, we have

 i = 3 7 ξ R 9                  (4)

 Now call the voltage across the two resistors on top V . Since the voltage across the two resistors in the bottom is iR, we have

 ξ = V + iR                      (5)

Using this and equation (4), we have 

V = 4 3 iR (6)

 Now that we have this voltage, we can compute the current through the resistor in the top left as 

itopleft = V 2R = 2 3 i (7) 

Finally, defining the current through the ammeter iA to be positive going to the left, current conservation tells us that

 iA + itopleft = i (8)

 and we combine this with equation (7) gives

 iA = 1 7 ξ R