Current coming from the battery and ammeter reading are

All three 4Ω resistors are in parallel .Assuming ammeter is ideal :
Current in all resistors will be divided and current will be $\frac{\mathrm{i}}{3}$

Effective resistance =$4+\frac{4}{3}=\frac{16}{3}\mathrm{\Omega }$

So, $\mathrm{i}=\frac{2}{{\displaystyle \frac{16}{3}}}=\frac{3}{8}\mathrm{A}$

Reading in ammeter = $\frac{1}{8}\mathrm{A}$