Here, we will use the Newton's law of cooling.

We know, $${{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)$$ .... (1)

where, T$_1$ and T$_2$ are initial and final temperatures respectively.

$\mathrm{T_s}$ = surrounding temperature

$\Delta t$ = time taken

$k$ = positive constant

Given,

Ist case :

$${T_1} = 90^\circ C$$

$${T_2} = 80^\circ C$$

$$\Delta t = t$$, $${T_s} = 20^\circ C$$

IInd case :

$${T_1} = 80^\circ C$$

$${T_2} = 60^\circ C$$

$${T_s} = 20^\circ C$$, $$\Delta t = ?$$

Now, for Ist case,

$${{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)$$ ...... (from (1))

$$ \Rightarrow {{10} \over t} = k\left( {85 - 20} \right)$$

$$ \Rightarrow {{10} \over t} = 65k$$ .... (2)

For IInd case,

$${{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)$$ .... (From (1))

$$ \Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)$$

$$ \Rightarrow {{20} \over {\Delta t}} = 50k$$ .... (3)

Now, by (2) $\div$ (3),

$${{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}$$

$$ \Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}$$

$$ \Rightarrow \Delta t = {{13} \over 5}t$$