Given data for 10 observations

The student originally computed:

• Mean $= 5.5$
• $\sum_{i=1}^{10} x_i^2 = 371$

Since there are $10$ observations, the original sum is $\sum x_i = 10 \times 5.5 = 55.$

Correct the mistaken entries

Two values were taken incorrectly as $4$ and $5$ instead of the correct values $6$ and $8$.

So, the corrected sum becomes $\sum x_i = 55 - (4+5) + (6+8) = 55 - 9 + 14 = 60.$

Hence the corrected mean is $\bar{x} = \frac{60}{10} = 6.$

Correct the sum of squares

Original: $\sum x_i^2 = 371.$

Replace $4^2$ and $5^2$ by $6^2$ and $8^2$: $\sum x_i^2 = 371 - (4^2+5^2) + (6^2+8^2).$

$= 371 - (16+25) + (36+64)$ $= 371 - 41 + 100 = 430.$

Use the variance formula

Variance of $n$ observations is $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2.$

Here, $\sigma^2 = \frac{430}{10} - 6^2 = 43 - 36 = 7.$

Match with the options

The variance of the corrected data is $\boxed{7}.$

Therefore, the correct option is B.