To find the maximum and minimum possible average weight, we need to find the maximum and minimum possible weights of students $a$ and $e.$
The sum of $a, b, c,$ and d is $160 \mathrm{~kg}$, and the sum of $b, c, d,$ and $e$ is $180 \mathrm{~kg}$. So, $e$ is $20$ more than $a.$
To find the highest possible value for the average weight, $e$ should have the highest possible value. This happens when $e$ is $20$ higher than the highest possible value for $a,$ which is $40 \mathrm{~kg}\;($all the first $4$ weights are equal to $40 \mathrm{~kg}).$ So, the highest possible average weight is $(160+60) / 5=44 \mathrm{~kg}$.
This will be the case when the weights are $40 \mathrm{kgs}, 40 \mathrm{kgs}, 40 \mathrm{kgs}, 40 \mathrm{kgs}$, and 60 kgs.
Conversely, the least possible value for the average weight occurs when a is the least. This happens when e is the least too $($since $a$ is $20$ less than $e).$
The least possible value for $e$ is $45 \mathrm{~kg}$. So, the least possible value for $a$ would-be $25 \mathrm{~kg}$. The least possible average weight is $(180+25) / 5=41 \mathrm{~kg}$.
This will be the case when the weights are $25 \mathrm{kgs}, 45 \mathrm{kgs}, 45 \mathrm{kgs}, 45 \mathrm{kgs}$, and $45 \;\text{kgs}.$
Therefore, the difference between the maximum and minimum possible average weight overall is $44 \mathrm{~kg}-41 \mathrm{~kg}=3 \mathrm{~kg}$.
Hence, the answer is$\text{ “3 kgs"}.$ Option C is the correct answer.