A cup of coffee cools from $90^\circ C$ to $80^\circ C$ in $t$ minutes, when the room temperature is $20^\circ C$. The time taken by a similar cup of coffee to cool from $80^\circ C$ to $60^\circ C$ at a room temperature same at $20^\circ C$ is:

Newton's law of cooling (average form): $\frac{T_1 - T_2}{t} = K\left[\frac{T_1 + T_2}{2} - T_0\right]$ First cup: $\frac{90 - 80}{t} = K\left[\frac{90 + 80}{2} - 20\right] = 65K$ ...(1) Second cup: $\frac{80 - 60}{t'} = K\left[\frac{80 + 60}{2} - 20\right] = 50K$ ...(2) Dividing (1) by (2): $\frac{t'}{2t} = \frac{65}{50} \Rightarrow t' = \frac{13}{5}t$