We will use the inclusion-exclusion principle. Brilliant, AoPS.
$A\cup BC\cup CDE\cup EF = A+ BC +CDE+ EF -(A\cap(BC))-(A\cap(CDE)) -(A\cap(EF))-((BC)\cap(CDE))-((BC)\cap(EF))-((CDE)\cap(EF))$
$+(A\cap(BC)\cap (CDE))+((BC)\cap(CDE)\cap (EF))+((A)\cap(CDE)\cap (EF))+((A)\cap(BC)\cap (EF))$
$-(A\cap (BC)\cap (CDE)\cap (EF))$
here $A=2^7$, no. of superkeys have A attribute
$BC=2^6$, no. of superkeys have BC attributes
$CDE=2^5$, no. of superkeys have CDE attributes
$EF=2^6$, no. of superkeys have EFattributes
$A\cap BC=2^5$, no. of superkeys have ABC attributes
$BC\cap CDE=2^4$, no. of superkeys have BC attributes
$A\cap CDE = 2^4$, no. of superkeys have ACDE attributes
$A\cap EF = 2^5$, no. of superkeys have AEF attributes
$BC\cap EF=2^4$, no. of superkeys have BCEFattributes
$CDE \cap EF=2^4$, no. of superkeys have CDEF attributes
$A\cap(BC)\cap (CDE)=2^3$, no. of superkeys have ABCDE attributes
$BC\cap CDE \cap EF=2^3$, no. of superkeys have BCDE attributes
$A\cap BC\cap EF=2^3$, no. of superkeys have ABCEF attributes
$A\cap (BC)\cap (CDE)\cap EF=2^2$, no. of superkeys have ABCDEF attributes
Chip in the values in the equation and we will get no. of superkeys =188
