Let the originally computed mean and standard deviation for $n=15$ observations be: $\bar{x}=12, \qquad s=3$

From the mean, $\sum x_i = n\bar{x} = 15\times 12 = 180$

Using variance relation: $s^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ Since $s=3$, we have $s^2=9$. Hence, $9 = \frac{\sum x_i^2}{15} - 12^2$ $9 = \frac{\sum x_i^2}{15} - 144$ $\frac{\sum x_i^2}{15} = 153$ $\sum x_i^2 = 15\times 153 = 2295$

One observation was taken as $10$ instead of the correct value $12$.

Therefore, the corrected sum is: $\sum y_i = 180 - 10 + 12 = 182$ So the correct mean is: $\mu = \frac{182}{15}$

Correct the sum of squares: $\sum y_i^2 = 2295 - 10^2 + 12^2 = 2295 - 100 + 144 = 2339$

For the corrected data, $\sigma^2 = \frac{\sum y_i^2}{15} - \mu^2$

Therefore, $\mu + \mu^2 + \sigma^2 = \mu + \frac{\sum y_i^2}{15}$ because $\mu^2 + \sigma^2 = \frac{\sum y_i^2}{15}$.

Multiply by $15$: $15(\mu+\mu^2+\sigma^2)=15\mu + \sum y_i^2$

Now, $15\mu = 182$ and $\sum y_i^2 = 2339$

Hence, $15(\mu+\mu^2+\sigma^2)=182+2339=2521$

Final answer: $\boxed{2521}$