The dust here means the average smallest amount that can be spent. One bitcoin sat is 0.00000001 but whether it can be spent is another question.

Define the followings:

• D average tx size 300 bytes
• B average block size
• R block reward 12.5
• F average block fee 1.9977
• $s_g$ mining power share of the mining pool in the global mining power
• $s_p$ mining power share of a miner in the mining pool
• d dust
• E the miner's mining power
• W the global mining power

Then, the tx fee of one typical transaction is $d=\frac{F}{B} D$. To make sense of 1 sat as dust, it means $10^{-8}=\frac{1.9977}{B} 300$, aka B is about 60 gigabytes. Should B be not that large, amount less then d is on average not spendable. For example if B is only 1 megabytes, the dust is 0.00059931, 5 promised digits are lost.

For the miner, he has probability $s_g$ to get a share of block reward and block fee which is $s_p (R+F)$ and it must be larger than $d=\frac{F}{B} D$. In other words, $s_p > \frac{F}{R+F} \frac{D}{B}$. Therefore the larger the block size, the more feasible a small miner can join the mining game.

Also, $E=s_g s_p W$. It follows $E=s_g s_p W > s_g \frac{F}{R+F} \frac{D}{B} W$. A usb-c mining chip miner has 250 watt, it follows this usb-c miner can play the mining game as long as $W<\frac{B}{D}\frac{R+F}{F}\frac{E}{s_g}$. Suppose a mining pool business is interesting to miners only if its mining power share is higher than 0.01, put all numbers together and it follows this usb-c miner can play the mining game until the global mining power is beyond $3.6 \times 10^{13}$ watt.