1)When A is candidate Key
then any combination from BCDEF along with A form Super Key.
Therefore ,total super key = $2^{5}$
2)When BC is candidate Key
In this case any combination from DEF along with BC form Super Key.
A is not considered in the case because all the combination of BC with A is counted in first case.
Ex={ABC,ABCD,....ABCDEF} all are counted in case 1.
Therefore ,total super key = $2^{3}$ (All combination from DEF)
3)When CD is candidate Key
In this case any combination from BEF along with CD form Super Key.
A is not considered (Similar logic as case B).
Along with we have also find the keys which are counted in case B.
{CDB,CDBE,CDBF,CDBEF} all these values are counted in case B,because all have BC(candidate key of case B).
So,these keys are counted twice because all have CD(candidate key )
Therefore ,total super key = $2^{3}$ (All combination from BEF)-$2^{2}$(ABCD as KEY)
total key = $2^{5}+2^{3}+2^{3}-2^{2}$