1. Given incorrect data statistics

For $n=20$ observations:

• Mean $=10$
• Standard deviation $=2$

Using the standard formulas: $\bar{x} = \frac{\sum x_i}{n}, \qquad \sigma = \sqrt{\frac{\sum x_i^2}{n} - \bar{x}^2}$

So, $\sum x_i = n\bar{x} = 20\cdot 10 = 200$

Also, $\sigma^2 = 4 = \frac{\sum x_i^2}{20} - 10^2$ $4 = \frac{\sum x_i^2}{20} - 100$ $\frac{\sum x_i^2}{20} = 104$ $\sum x_i^2 = 2080$

2. Correct the mistaken observation

One observation was taken as $8$ instead of $12$.

Corrected sum

$\sum x_i' = 200 - 8 + 12 = 204$

Hence corrected mean is $\bar{x}' = \frac{204}{20} = 10.2$

Corrected sum of squares

$\sum {x_i'}^2 = 2080 - 8^2 + 12^2$ $= 2080 - 64 + 144$ $= 2160$

3. Find corrected variance

$\sigma'^2 = \frac{\sum {x_i'}^2}{20} - (\bar{x}')^2$ $= \frac{2160}{20} - (10.2)^2$ $= 108 - 104.04$ $= 3.96$

Therefore, the corrected standard deviation is $\sigma' = \sqrt{3.96}$

4. Match with options

The correct option is:

$\boxed{\text{B: } \sqrt{3.96}}$