It is given in the question that:

Temperature of the surroundings = ${\mathrm{T}}_{\mathrm{o}}$ = 20 °C

According to Newton’s law of cooling,

$$\begin{gathered} -\frac{\mathrm{dt}}{\mathrm{dt}}=\mathrm{K}(\mathrm{T}-{\mathrm{T}}_0) \\ \frac{\mathrm{dT}}{\mathrm{K}(\mathrm{T}-{\mathrm{T}}_0)}=-\mathrm{Kdt} .....\left(\mathrm{i}\right) \end{gathered}$$

Where,

The temperature of the body = T

K = constant

When the temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Integrating equation (i), 

${\int }_{50}^{80}\frac{\mathrm{dT}}{\mathrm{K}\left(\mathrm{T}-{\mathrm{T}}_0\right)}=-{\int }_0^{300}\mathrm{Kdt}$

$$\begin{gathered} \Rightarrow {\left[{\log }_{\mathrm{e}}(\mathrm{T}-{\mathrm{T}}_0)\right]}_{50}^{80}=-\mathrm{K}{\left[\mathrm{t}\right]}_0^{300} \\ \Rightarrow \frac{2.3026}{\mathrm{K}}{\log }_{10}\left(\frac{80-20}{50-20}\right)=-300 \\ \Rightarrow \frac{2.3026}{\mathrm{K}}{\log }_{10}2=-300 \\ \Rightarrow \mathrm{K}=\frac{-2.3026}{300}{\log }_{10}2 \end{gathered}$$

Let the temperature of the body falls from 60°C to 30°C in time = t.
Hence,

$$\begin{gathered} \frac{2.3026}{\mathrm{K}}{\log }_{10}\frac{60-20}{30-20}=-\mathrm{t} \\ \frac{-2.3026}{\mathrm{t}}{\log }_{10}4=\mathrm{K} ....\left(\mathrm{iii}\right) \end{gathered}$$

Equating equations (ii) and (iii),

$$\begin{gathered} \Rightarrow \frac{-2.3026}{\mathrm{t}}{\log }_{10}4=\frac{-2.3026}{300}{\log }_{10}2 \\ \Rightarrow \mathrm{t}=300\times 2=600 \mathrm{s}=10 \min \end{gathered}$$

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.