The current $$I$$ in the given circuit will be

The four edge-resistors form a Wheatstone-bridge arrangement. Let the resistances be $$R_{AB}=2 \,\Omega$$, $$R_{BC}=2 \,\Omega$$, $$R_{AD}=2 \,\Omega$$ and $$R_{DC}=2 \,\Omega$$. The fifth resistor $$R_{BD}=5 \,\Omega$$ is connected between the bridge nodes $$B$$ and $$D$$.

For a Wheatstone bridge, the branch $$BD$$ carries zero current when the ratio of the two resistors in the left arm equals the ratio in the right arm, i.e. when $$\frac{R_{AB}}{R_{BC}}=\frac{R_{AD}}{R_{DC}}$$. Substituting the given values, $$\frac{2}{2}=1=\frac{2}{2}$$, so the bridge is balanced. Hence no current flows through the $$5 \,\Omega$$ resistor and we can remove it from further calculations.

With the central branch ignored, the circuit reduces to two simple series arms placed in parallel between the terminals of the battery:

Left arm: $$R_{AB}+R_{BC}=2+2=4 \,\Omega$$
Right arm: $$R_{AD}+R_{DC}=2+2=4 \,\Omega$$

These two equal resistances are in parallel, so the equivalent resistance between the battery terminals is

$$\frac{1}{R_{\text{eq}}} = \frac{1}{4}+\frac{1}{4}= \frac{2}{4} = \frac{1}{2}$$
$$\Rightarrow \; R_{\text{eq}} = 2 \,\Omega$$

The emf of the battery supplied in the circuit diagram is $$E = 20 \text{ V}$$. Using Ohm’s law, the current delivered by the battery is

$$I = \frac{E}{R_{\text{eq}}} = \frac{20 \text{ V}}{2 \,\Omega} = 10 \text{ A}$$

Hence the current in the circuit is $$10 \text{ A}$$.

Option A is correct.