The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _____.

Given: 10 students, mean = 50, SD = 12. Two marks 20 and 25 were wrongly read as 45 and 50.

Wrong $$\sum x = 50 \times 10 = 500$$. Wrong $$\sum x^2 = n(\sigma^2 + \bar{x}^2) = 10(144 + 2500) = 26440$$.

Correct $$\sum x = 500 - 45 - 50 + 20 + 25 = 450$$. Correct mean = 45.

Correct $$\sum x^2 = 26440 - 45^2 - 50^2 + 20^2 + 25^2 = 26440 - 2025 - 2500 + 400 + 625 = 22940$$.

Correct variance = $$\frac{\sum x^2}{n} - \bar{x}^2 = \frac{22940}{10} - 2025 = 2294 - 2025 = 269$$.

The answer is $$\boxed{269}$$.