Given, cup of tea cools from $65.5^{\circ} \mathrm{C}$ to $62.5^{\circ} \mathrm{C}$ in 1 minute.
Room temperature $=22.5^{\circ} \mathrm{C}$
We know that $\frac{d T}{d t}=k\left(\theta-\theta_{0}\right) \rightarrow(1)$
For 1st case $: \frac{d T}{d t}=\frac{65.5-62.5}{1 \mathrm{~min}}=\frac{3^{\circ} \mathrm{C}}{1}$
Also,
$\theta-\theta_{0}=\left(\frac{65.5+62.5}{2}\right)-22.5=64-22.5$
$\theta-\theta_{0}=41.5^{\circ} \mathrm{C}$
Now, substituting in Eq. (1), we get
$3=k(41.5)$
$\Rightarrow k=\frac{3}{41.5} \rightarrow(2)$
For 2 nd case $: \frac{d T}{d t}=\frac{46.5-40.5}{t}=\frac{6^{\circ} \mathrm{C}}{t}$
Also,
$\theta-\theta_{0}=\left(\frac{46.5+40.5}{2}\right)-22.5=43.5-22.5$
$\Rightarrow \theta-\theta_{0}=21^{\circ} \mathrm{C}$
Again, substituting in Eq. (1), we get
$\frac{6}{t}=k \times 21$
Substitute value of k from Eq. (2), we get
$\frac{6}{t}=\frac{3}{41.5} \times 21$
$\Rightarrow t=\frac{6 \times 41.5}{3 \times 21}=3.9523$