First, find candidate key and they are A,C,E
According to the principle of inclusion and exclusion,
∣A ⋃ C ⋃ E∣ = |A| + |C| + |E| - |A ⋂ C| - |A ⋂ E| - | C ⋂ E| + |A ⋂ C ⋂ E|
= No. of super key using A + No. of superkey using C + No. of superkey using E - No. of superkeys using A and C - No. of super keys using A and E - No. of superkeys using C and E
+ No. of super keys using A,C,E
= $2^{n-1}$ + $2^{n-1}$ + $2^{n-1}$ - $2^{n-2}$ - $2^{n-2}$ - $2^{n-2}$ + $2^{n-3}$ , Here n = 5 (No. of attributes in a relation)
= $2^4$ + $2^4$ + $2^4$ - $2^3$ - $2^3$ - $2^3$ + $2^2$
= 28
So the answer is : 28
Note: In general, if we have ‘N’ attributes with one candidate key then the number of possible superkeys are $2^{n-1}$.