According to Newton's law of cooling, we have:

$-\frac{d T}{d t}=K\left(T-T_{0}\right)$

$\frac{d T}{K\left(T-T_{0}\right)}=-K d t…(i)$

Where,

Temperature of the body $=T$

Temperature of the surroundings $=T_{0}=20^{\circ} C$

$K$ is a constant

Temperature of the body falls from $80^{\circ} C$ to $50^{\circ} C$ in time, $t=5 min =300 s$

Integrating equation $(i),$ we get

$\int_{00}^{80} \frac{d T}{K\left(T-T_{0}\right)}=-\int_{0}^{300} K d t$

$\left[\log _{e}\left(T-T_{0}\right)\right]_{s_{0}}^{80}=-K[t]_{0}^{300}$

$\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}=-300$

$\frac{2.3026}{K} \log _{10} 2=-300$

$\frac{-2.3026}{300} \log _{10} 2=K\dots (ii)$

The temperature of the body falls from $60^{\circ} C$ to $30^{\circ} C$ in time $=t'$

Hence, we get:

$\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}=-t$

$\frac{-2.3026}{t} \log _{10} 4=K\dots (iii)$

Equating equations ( $i i$ ) and (iii), we get:

$\frac{-2.3026}{t} \log _{10} 4=\frac{-2.3026}{300} \log _{10} 2$

$t=300 \times 2=600 s =10 min$

Therefore, the time taken to cool the body from $60\,^{\circ} C$ to $30\,^{\circ} C$ is $10$ minutes.