Resistance of the ammeter is

$R_{A}=\frac{(480\, \Omega)(20\, \Omega)}{(480\, \Omega+20\, \Omega)}=19.2\, \Omega$

(As $480 \,\Omega$ and $20\, \Omega$ are in parallel)

As ammeter is in series with $40.8\, \Omega$,

$\therefore \quad$ Total resistance of the circuit is

$R=40.8\, \Omega+R_{A}=40.8\, \Omega+19.2\, \Omega=60\, \Omega$

By Ohm's law,

Current in the circuit is

$I=\frac{V}{R}=\frac{30\, \mathrm{V}}{60\, \Omega}=\frac{1}{2} \mathrm{A}=0.5\, \mathrm{A}$

Thus the reading in the ammeter will be $0.5\, \mathrm{A}$.