Answer : Option D

Explanation :

Let the period be n years.
Amount after n years = Rs.20000 + Rs.3328 = Rs. 23328
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 23328\\\\ 20000\left(1 + \dfrac{8}{100}\right)^\text{n} = 23328\\\\ 20000\left(\dfrac{108}{100}\right)^\text{n} = 23328\\\\ \left(\dfrac{108}{100}\right)^\text{n} = \dfrac{23328}{20000} = \dfrac{11664}{10000} = \left(\dfrac{108}{100}\right)^2\\\\ \text{n = 2 years}$MF#%