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

We have

$$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$$

As per given information correct $$\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$$

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

Also

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

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

$$\begin{aligned} & =15\left(\mu+\mu^2+\sigma^2\right) \\ & =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\ & =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\ & =2521 \end{aligned}$$