Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$

We have

$\mu^{\prime}=\frac{\Sigma x_i}{15}=12 \Rightarrow \Sigma x_i=180$

As per given information correct $\Sigma x_i=180-10+12$

$\Rightarrow \mu(\text { correct mean })=\frac{182}{15}$

Also

$ \sigma^{\prime}=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295 $

$\text { Correct } \Sigma \mathrm{x}_{\mathrm{i}}^2=2295-100+144=2339 $

$ \sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$

Required value

$ =15\left(\mu+\mu^2+\sigma^2\right) $

$ =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) $

$ =15\left(\frac{182}{15}+\frac{2339}{15}\right) $

$ =2521$