1. Given data for the incorrectly recorded observations

• Number of observations: $n=10$
• Mean: $\bar{x}=20$
• Standard deviation: $8$

So the variance of the incorrect data is $\sigma^2 = 8^2 = 64.$

2. Find the sum and sum of squares of the incorrect data

Using $\bar{x} = \frac{\sum x_i}{n},$ we get $\sum x_i = n\bar{x} = 10\cdot 20 = 200.$

Also, $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2.$ So, $64 = \frac{\sum x_i^2}{10} - 20^2.$ $64 = \frac{\sum x_i^2}{10} - 400$ $\frac{\sum x_i^2}{10} = 464$ $\sum x_i^2 = 4640.$

3. Correct the wrong observation

One observation was recorded as $50$ instead of $40$.

So the correct sum is $\sum x_i^{(\text{correct})} = 200 - 50 + 40 = 190.$

And the correct sum of squares is $\sum (x_i^2)^{(\text{correct})} = 4640 - 50^2 + 40^2.$ $= 4640 - 2500 + 1600 = 3740.$

4. Find the correct mean

$\bar{x}_{\text{correct}} = \frac{190}{10} = 19.$

5. Find the correct variance

Using $\sigma_{\text{correct}}^2 = \frac{\sum x_i^2}{n} - \bar{x}_{\text{correct}}^2,$ we get $\sigma_{\text{correct}}^2 = \frac{3740}{10} - 19^2.$ $= 374 - 361 = 13.$

6. Conclusion

The correct variance is $\boxed{13}.$

So the correct option is C.