Let's say the Temperature of body at time $t$ be $T_{t} $, which means $T_{0} = 30.$
Now let the time at which temperature is $5$ be $t_1,$ which means $T_{t_1} = 5$
and the time at which temperature is $1$ be $t_2$ , which means $T_{t_2} = 1.$
Question Asks us to find $\dfrac{t_2}{t_1}$.
Now,$\text{ Temperature decrease(D) at time} t\propto \text{Body Temperature} - \text{Bath Temperature}$
$D\propto T_{t} \;\{\text{As Bath Temp. is 0 & Body Temp. is Tt}\}$
$D = k\times T_{t}$ {where $k$ is proportionality constant}
now, it gives $T_{t+1} = T_{t} - D = T_{t} - k\times T_{t} = (1-k) T_{t}$
Now,$T_0 = 30$
$T_1 = 30 (1-k)$
$T_2 = 30(1-k)^{2}$
$T_{t_1} = 30(1-k)^{t_1}$ & $T_{t_2} = 30(1-k)^{t_2}$
$\Rightarrow 30(1-k)^{t_1} = 5$
$\Rightarrow t_1\times \log (1-k) = \log \left(\dfrac{5}{30}\right) = \log \left(\dfrac{1}{6}\right)$
Similarly $\Rightarrow t_{2}\times \log (1-k) = \log \left(\dfrac{1}{30}\right)$
$\dfrac{t_{2}}{ t_{1}} =\dfrac{\log \left(\dfrac{1}{30}\right)}{\log \left(\dfrac{1}{6}\right)}$
$=\dfrac{\log (30-1)}{\log (6-1)} =\dfrac{\log(30)} {\log (6)}=\log_{6} 30$
$=\log_{6} (6\times 5) =\log_{6} 6 +\log_{6} 5 =1 +\log_{6} 5$
which is OPTION (D)