The thermal resistance of rod 1 will be R1=KAl1=KA4 and rod 2 will be R2=2KAl2=2KA2.5.
Since they are connected in series, Req=R1+R2=2KA10.5.
If equivalent thermal conductivity is Keq, then Req=KeqAl1+l2=KeqA6.5. Therefore,
2KA10.5=KeqA6.5⇒Keq=2126K=(1+215)K
Hence, α=21.