(a,c) 

$\left(a\right)\hspace{0.17em}{H}_{3}C—CH—Br C{H}_3$
$| |$
$C{H}_3 \left(b\right) H_3—C—C{H}_{2}Br$
$|$
$C{H}_3$
$\left(c\right) H_3—CH—C{H}_2—C{H}_{3 } Br$
$| |$
$Br \left(d\right) H_{3}C—C—C{H}_{2}C{H}_3$
$|$
$C{H}_3$

Secondary bromides are those compounds in which a-carbon (ie. carbon bonded to bromine) is further bonded to two alky! groups. In compound (a) and (c} a-carbon is bonded to two alkyl groups thats why it is secondary bromide but in compound (b) it is bonded to one alky group and it is primary bromide, In compound (c) it is bonded to three alkyl groups and itis tertiary alkyl halide.