Correct Answer - A::C
`K(cu)=390W//m-^(@)C`
`K(st)=46W//m-^(@)C`
Now since they are in series connection the heat passed throogh the cross section is the same
so, `Q_1=Q_2`
`or (K_(cu)xxAxx(theta-0))/l`
`=(K_(ST)xxAxx(100-theta))/l`
`implies 390(theta-0)=46xx100-46 theta`
`implies 436 theta=4600`
`theta=4600/436`
`-10.55=10.6^(@)C.`