Because the question is flawed or inadequate in phrasing (asking
only for 'stereoisomers' implies no structural isomers are formed,
a false assumption), hence there is no satisfactory answer. It may
well be that all the available MCQ choices are wrong in the Prelim
paper.
Notice that there are two possible structural isomers (the
double bond at the bottom of the ring applies to both
structural isomers, and isn't capable of cis-trans or E-Z, so
ignore it in your counting), both of which have diastereomeric
("geometric") isomers - one is capable of cis/trans and E/Z, the
other capable of E/Z only (many JCs wrongly teach students to
artificially force cis/trans isomeric categorization for
many alkenes that can actually only have E/Z. Technically,
cis-trans isomers exist only when an alkene has two substituents
and two hydrogens attached to the sp2 hybridized carbons).
After dehydration by excess conc H2SO4, only one of the two
abovementioned resulting structural isomers have a
single stereocenter / chiral center.
So if the question was reiterated or rephrased to "how many
isomers in total (include both structural and stereoisomers) may be
generated?", then the correct answer would be
structural isomer #1 - E - rectus
structural isomer #1 - E - sinister
structural isomer #1 - Z - rectus
structural isomer #1 - Z - sinister
structural isomer #2 - E
structural isomer #2 - Z
Alternatively,
structural isomer #1 - cis - rectus
structural isomer #1 - cis - sinister
structural isomer #1 - trans - rectus
structural isomer #1 - trans - sinister
structural isomer #2 - E
structural isomer #2 - Z
Or,
structural isomer #1 - cis - dextrorotatory
structural isomer #1 - cis - levorotatory
structural isomer #1 - trans - dextrorotatory
structural isomer #1 - trans - levorotatory
structural isomer #2 - E
structural isomer #2 - Z
Or,
structural isomer #1 - E - dextrorotatory
structural isomer #1 - E - levorotatory
structural isomer #1 - Z - dextrorotatory
structural isomer #1 - Z - levorotatory
structural isomer #2 - E
structural isomer #2 - Z