The value of net resistance of the network as shown in the given figure  is:

First, note the potentials:
left node = −6 V, right node = −8 V

So left is at higher potential → current tends to flow left → right.

Now check diodes:

• middle branch (10 Ω): diode allows left → right → forward biased 
• bottom branch (5 Ω): diode allows right → left → reverse biased (no current)

So active branches:

• top: 15 Ω
• middle: 10 Ω
(bottom 5 Ω is inactive)

Now these two are in parallel:

$$R_{eq}=15∥10=\frac{\left(15\times10\right)}{15+10}=\frac{150}{25}=6Ω$$