The current $$I_1$$ (in A) flowing through 1$$\Omega$$ resistor in the following circuit is:

Equivalent resistance of the parallel $$1\ \Omega$$ pair: $$R_{p1} = \frac{1 \times 1}{1 + 1} = 0.5\ \Omega$$

Total resistance of the top branch connected across the battery: $$R_{\text{top}} = R_{p1} + 2 = 0.5 + 2 = 2.5\ \Omega$$

Total current entering the top branch from the $$1\text{ V}$$ source: $$I_{\text{top}} = \frac{V}{R_{\text{top}}} = \frac{1}{2.5} = 0.4\text{ A}$$

Current $$I_1$$ splitting equally into the identical $$1\ \Omega$$ parallel branch:

$$I_1 = \frac{I_{\text{top}}}{2} = \frac{0.4}{2} = 0.2\text{ A}$$