From Newton's Law of cooling,
ΔtΔT=k(T−Ts)
where,
⇒tT2−T1−k(2T1+T2−Ts)Ts
Where, T1= Initial temperature of body
T2= Final temperature of body
Ts= surrounding temperature
t= time taken
For case1,
t90−80=k(290+80−20)
⇒t10=65k
⇒k=65t10…(i)
For case 2,
t′80−60=k(280+60−20)
⇒t′20=50k
Put the value k from eq.(i),
⇒t′20=50×65t10
⇒t′=50020×65t
⇒t′=513t