Answer
5.48 mA.
Work Step by Step
The total resistance in the circuit, with the ammeter in place, is $R_{eq}=53\Omega+720\Omega+480\Omega=1253\Omega$.
Solve for the voltage of the (ideal) battery by using Ohm’s law.
$$V_{battery}=IR_{eq}=(5.25\times10^{-3}A)(1253\Omega)=6.578V$$
Now remove the ammeter, and assume that the battery voltage stays the same. The equivalent resistance is now $R_{eq}=53\Omega+720\Omega+480\Omega$. Find the new current from Ohm’s law.
$$I=\frac{V_{battery}}{R_{eq}}=\frac{6.578V}{1200\Omega}=5.48\times10^{-3}A$$
