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

The two 10 Ω resistors inside are connected in parallel between the left node and middle node.

The 10 V battery has positive terminal on the right, so:

$$V_B-V_A=10\text{ V}$$

Thus voltage across each parallel resistor is 10 V.

Current through each:

$$I_1=I_2=\frac{10}{10}=1\text{ A}$$

(direction shown right to left, matching arrows)

Now consider the outer loop.

The 20 V battery has positive terminal on the left, so moving left to right across it gives a drop of 20 V20\text{ V}20 V.

Taking left node as 0 V:

• Middle node $$=+10\text{ V}$$
• Right node$$=10-20=-10\text{ V}$$

Voltage across bottom 10 Ω10\,\Omega10Ω resistor:

$$0-(-10)=10\text{ V}$$

Hence,

$$I_3=\frac{10}{10}=1\text{ A}$$

Now,