Determine the current recorded by the ammeter for the circuit...

The given circuit has a horizontal axis of symmetry.

Due to this symmetry, the potentials at the corresponding nodes in the upper and lower branches are equal. Therefore, the potential difference across each of the $50\ \Omega$ resistors is zero, and no current flows through them.

The $50\ \Omega$ resistors can be removed from the circuit.

The simplified circuit consists of two parallel branches, each with three $10\ \Omega$ resistors in series.

Resistance of the upper branch $= 10 + 10 + 10 = 30\ \Omega$.

Resistance of the lower branch $= 10 + 10 + 10 = 30\ \Omega$.

Equivalent resistance of the circuit, $R_{eq} = \dfrac{30 \times 30}{30 + 30} = 15\ \Omega$.

The current recorded by the ammeter is $I = \dfrac{V}{R_{eq}} = \dfrac{6}{15} = 0.4\text{ A}.