Degree of unsaturation of naphthalene Double Bond Equivalent/Degree of Unsaturation
  
  25. Degree of unsaturation in
  
  is
  
  (a) 6; (b) 7; (c) 8.


There are two rings and six double bonds in structure so degree of unsaturation should be 8, but it is not. Why?
You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct.
You can redraw the structure of naphthalene as such:



and make sure that there are five double bonds and two rings giving degree of unsaturation

$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$

You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there: 

$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$

You again arrive at $\text{DU} = 7$ using a simplified formula:

$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$