The value of current $$i_1$$ flowing from A to C in the circuit diagram is:

First, let us read the information that is visible in the circuit. Between the terminals A and C there are two resistors, one of $$2\; \Omega$$ (marked between A and some intermediate point B) and the other of $$4\; \Omega$$ (marked between the same intermediate point B and C). Downstream, after C, a third resistor of $$6\; \Omega$$ joins the return line back to the negative terminal of a single ideal source of $$12\; \text{V}$$ whose positive terminal is connected to A. In words, the three resistors of $$2\; \Omega, 4\; \Omega,$$ and $$6\; \Omega$$ are all in one line, so they are in series.

Whenever several resistors are in series, we add their resistances to get one equivalent resistance. Stating the formula for resistors in series,

$$R_{\text{eq}} \;=\; R_1 \;+\; R_2 \;+\; R_3 \;+\; \cdots$$

Here we have only three resistors, so

$$R_{\text{eq}} \;=\; 2\; \Omega \;+\; 4\; \Omega \;+\; 6\; \Omega \;=\; 12\; \Omega.$$

Because the entire string of resistors is connected directly across the $$12\; \text{V}$$ battery, this whole equivalent resistance of $$12\; \Omega$$ experiences the full $$12\; \text{V}$$ of the source.

Now we invoke Ohm’s Law, which in its basic form is stated as

$$I \;=\; \dfrac{V}{R}.$$

Substituting the values we have just extracted,

$$I \;=\; \dfrac{12\; \text{V}}{12\; \Omega} \;=\; 1\; \text{A}.$$

This current of $$1\; \text{A}$$ is the same through every series element, because in a series circuit the current has only one possible path. In particular, the branch from A to C carries exactly this current. The symbol $$i_1$$ defined in the question as the current flowing from A to C is therefore equal to $$1\; \text{A}.$$

Hence, the correct answer is Option D.