Решение

$\bar{v} = \sqrt{\frac{3kT}{m}}$$\Rightarrow m = \frac{3kT}{v^2}$ $m_0 \gg m;$ $m_0 \gg \frac{3kT}{v^2}$ $m_0 \gg 1.1 \cdot 10^{-9} \,кг$ $m_0 \sim 10^{-8} \, кг$ $\approx 0.01 \,мг$