Since the average of $a, b$ and $c$ is 16, then rac{a+b+c}{3}=16 and so $a+b+c=3 imes 16=48$. Since the average of $c, d$ and $e$ is 26, then rac{c+d+e}{3}=26 and so $c+d+e=3 imes 26=78$. Since the average of $a, b, c, d$, and $e$ is 20, then rac{a+b+c+d+e}{5}=20. Thus, $a+b+c+d+e=5 imes 20=100$. We note that $(a+b+c)+(c+d+e)=(a+b+c+d+e)+c$ and so $48+78=100+c$ which gives $c=126-100=26$.