Only the isomers of C5H10 (acyclic alkenes and cycloalkanes, counting stereoisomers separately wherever they exist) are considered.

Enumerate the constitutions and then the stereoisomers. Acyclic alkenes: pent-1-ene, pent-2-ene (the $E$ and $Z$ forms), 2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene, giving $1+2+1+1+1=6$ isomers, so (A) is wrong. Cycloalkanes: cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane and 1,2-dimethylcyclopropane. The last exists as the achiral cis (meso) form and the trans form as a pair of enantiomers, i.e. three isomers, so the cycloalkanes total $1+1+1+1+3=7$ and (D) is wrong. The overall total is $6+7=13$, so (B) is correct. The only optically active isomers are the two enantiomers of trans-1,2-dimethylcyclopropane, so the number is $2$ and (C) is wrong.