In the given circuit, the value of $$\left|\frac{I_1+I_3}{I_2}\right|$$ is:

Let us set the potential of the left vertical wire to $$0\text{ V}$$ (ground reference). Since all points on a continuous wire are at the same potential, the entire left side of the three resistors is at $$0\text{ V}$$.

Node B (Middle): $$V_B = 0\text{ V} + 10\text{ V} = 10\text{ V}$$

Node C (Far Right): $$V_C = V_B - 20\text{ V} = 10\text{ V} - 20\text{ V} = -10\text{ V}$$

$$I_1 = \frac{V_B - 0}{10} = \frac{10 - 0}{10} = 1\text{ A}$$

$$I_2 = \frac{V_B - 0}{10} = \frac{10 - 0}{10} = 1\text{ A}$$

$$I_3 = \frac{0 - V_C}{10} = \frac{0 - (-10)}{10} = \frac{10}{10} = 1\text{ A}$$

$$\text{Value} = \left| \frac{I_1 + I_3}{I_2} \right|$$

$$\text{Value} = \left| \frac{1 + 1}{1} \right| = \left| \frac{2}{1} \right| = 2$$