The normality of 10% (w/v) acetic acid is:

The correct option is C 1.67 N
10% w/v means 10 g of acetic acid is present in 100 mL of solution.
Moles of acetic acid = $\frac{\text{given mass}}{\text{molar mass}}=\frac{10}{60}=0.167$

$\text{Molarity}=\frac{\text{moles of solute}}{\text{total volume of solution (mL)}}\times 1000=\frac{0.167}{100}\times 1000=1.67M$
Since the n-factor for acetic acid is 1, normality = 1.67 N