Answer
The ammeter reads $~~0.45~A$
Work Step by Step
We can find the equivalent resistance of $R_2$ and $R_3$:
$\frac{1}{R_{23}} = \frac{1}{4.0~\Omega}+ \frac{1}{6.0~\Omega}$
$R_{23} = 2.4~\Omega$
Then the equivalent resistance in the circuit is $2.4~\Omega+ 2.0~\Omega = 4.4~\Omega$
We can find the total current in the circuit:
$i = \frac{\mathscr{E}}{4.4~\Omega} = \frac{5.0~V}{4.4~\Omega} = 1.14~A$
We can find the potential difference across $R_1$:
$\Delta V = (1.14~A)(2.0~\Omega) = 2.28~V$
Then the potential difference across $R_3$ is $~~5.0~V-2.28~V = 2.72~V$
We can find the current through $R_3$:
$i_3 = \frac{2.72~V}{6.0~\Omega} = 0.45~A$
Since this is the same current through the ammeter, the ammeter reads $~~0.45~A$
