If the sum of the first ten terms of the series $\left(1\frac{3}{5}\right)^{2}+\left(2\frac{2}{5}\right)^{2}+\left(3\frac{1}{5}\right)^{2}+4^{2}+\left(4\frac{4}{5}\right)^{2}+\ldots$, is $\frac{16}{5} m$, then $m$ is equal to:
(1) 102
(2) 101
(3) 100
(4) 99