Let,$G$ denote normalized order traffic load in the channel.
Given 20% slots are idle which implies frames will be successfully transmitted if we send in those 20% of idle slots.
Probability of successfull transmission =$P_{succ}$=0.2
we know , In slotted aloha probability of successfull transmission=$e^{-G}$
so ,
$e^{-G}=0.2$
$G=-\log_{e}0.2=1.609$
So, Normalized order traffic load=1.609.
Normalized Throughput =$G*P_{succ}=1.609*0.2=0.321$=32%
Why in slotted aloha probability of successfull transmission = $e^{-G}$ ?
derivation :- https://wiki.eecs.yorku.ca/course_archive/2009-10/W/3213/_media/cse3213_14_randaccess1_w2010.pdf