Answer : Option B

Explanation :

Let the sum be P and rate of interest be R% per annum.
Amount after 3 years = 2200
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 2200\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3 = 2200 \quad \color{#F00}{\text{--- ( 1)}}$MF#%
Amount after 6 years = 4400
$MF#%\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^\text{T} = 4400\\\\ \text{P}\left(1 + \dfrac{\text{R}}{100}\right)^6 = 4400 \quad \color{#F00}{\text{--- (2)}}$MF#%
$MF#%\color{#F00}{\text{(2)÷(1) =>}} \dfrac{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^6 }{\text{P}\left(1 + \dfrac{\text{R}}{100}\right)^3} = \dfrac{4400}{2200} = 2\\\\ => \left(1 + \dfrac{\text{R}}{100}\right)^3 = 2 \\\\ \color{#F00}{\text{(Substituting this value in equation 1) => }} \quad \text{P} \times 2 = 2200\\\\ \text{P} = \dfrac{2200}{2} = 1100$MF#%