has only the prime divisors  and .
Every divisor of  is therefore of the form ,
where  and  are elements of {0,1,2,3}.  Since there
are  choices for  and  choices for , there are 
divisors of .  That isn't necessary to know.  But
in the product of all  divisors of , 
or , and  or  occur exactly  times each.
Since  and  both occur  times each in the product
of all divisors, the product of the divisors must be
, choice a)

Edwin