The sum of first $n,n\ge 1$ terms of arithmetic progression with first term 2 and common difference $4,$ is

$S_n=\frac{n}{2}[4+(n-1\left)4\right]=2n^2$

So, the average of the first $n$ terms

$M_n=\frac{S_n}{n}=2n$

$\text{ Now, }\sum _{n=1}^{10} M_n=2\sum _{n=1}^{10} n$

$=2\times \left(\frac{10\times 11}{2}\right)=110$