Answer : Option C

Explanation :

Let the sum be P
The sum P becomes 3P in 4 years on compound interest
$MF#%3\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^4\\\\ \Rightarrow 3 = \left(1 + \dfrac{\text{R}}{100}\right)^4$MF#%
Let the sum P becomes 81P in n years
$MF#%81\text{P} = \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^n\\\\ \Rightarrow 81 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow (3)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^n \\\\ \Rightarrow \left(\left(1 + \dfrac{\text{R}}{100}\right)^4\right)^4 = \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \Rightarrow \left(1 + \dfrac{\text{R}}{100}\right)^{16}= \left(1 + \dfrac{\text{R}}{100}\right)^\text{n} \\\\ \text{n} = 16$MF#%
i.e, the sum will become 81 times in 16 years