Explanation:
All the five resistors are connected in parallel to 300 \(V\) battery.
\(\therefore \) Current through each resistor is
\(I = \frac{V}{R} = \frac{{300\,\,\,V}}{{1500\,\,\,\Omega }} = \frac{1}{5}\,A\)
Current flowing through the ammeter \(A\) is the sum of currents flowing through three the resistances to its right
\(3\,I\, = \,3\left( {\frac{1}{5}\,A} \right) = \frac{3}{5}\,A\)