B

Step 1: We start by using the mirror formula:
$$ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $$
where:
- $f$ is the focal length of the concave mirror,
- $u$ is the object distance (taken as negative for mirrors),
- $v$ is the image distance (taken as positive for real images).

Step 2: Initially, the object distance $u = -10$ cm and the image distance $v = 20$ cm.
Substituting these values in the formula:
$$ \frac{1}{f} = \frac{1}{-10} + \frac{1}{20} $$
$$ \frac{1}{f} = -\frac{1}{10} + \frac{1}{20} = -\frac{2}{20} + \frac{1}{20} = -\frac{1}{20} $$
Thus, the focal length $f = -20$ cm.

Step 3: Now, if the object is moved 0.1 cm towards the mirror, the new object distance is:
$$ u' = -10 + 0.1 = -9.9 ext{ cm} $$

Step 4: We will calculate the new image distance $v'$ using the mirror formula again:
$$ \frac{1}{f} = \frac{1}{u'} + \frac{1}{v'} $$
Substituting the values of $f$ and $u'$:
$$ -\frac{1}{20} = \frac{1}{-9.9} + \frac{1}{v'} $$

Step 5: Rearranging gives us:
$$ \frac{1}{v'} = -\frac{1}{20} - \frac{1}{-9.9} $$
$$ \frac{1}{v'} = -\frac{1}{20} + \frac{1}{9.9} $$
Finding a common denominator (1980):
$$ \frac{1}{20} = \frac{99}{1980}, \quad \frac{1}{9.9} = \frac{200}{1980} $$
Thus, substituting gives us:
$$ \frac{1}{v'} = -\frac{99}{1980} + \frac{200}{1980} = \frac{101}{1980} $$
Therefore, calculating $v'$:

$$ v' = \frac{1980}{101} \approx 19.62 ext{ cm} $$

Step 6: Finally, the shift in the image position ($\Delta v$) is given by:
$$ \Delta v = v - v' $$
Where $v = 20$ cm and $v' \approx 19.62$ cm:
$$ \Delta v = 20 - 19.62 = 0.38 ext{ cm} = 3.8 ext{ mm} $$

Therefore, the shift of image is 3.8 mm.