As the temperature differences are small, we can use Newton's law of cooling.

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)\frac{d\theta }{\theta -{\theta }_0}=-kdt$                                                                          ...(i)
Where k is a constant,$\theta$ is the temperature of the body at time t and ${\theta }_0$ =  ${16}^{\circ }C$ is the temperature of the surrounding. We have,

${\int }_{{40}^{0}c}^{{36}^{0}c}\frac{d\theta }{\theta -{\theta }_0}=-k\left(5min\right)In\frac{{36}^{0}c-{16}^{0}c}{{40}^{0}c-{16}^{0}c}=-k\left(5min\right)$

or, $k=-\frac{In\left(5/6\right)}{5min}$

or, If t be the time required for the temperature to fall from 36⁰C to 32⁰C then by (i),

${\int }_{{36}^{0}c}^{{32}^{0}c}\frac{d\theta }{\theta -{\theta }_0}=-ktIn\frac{{32}^{0}c-{16}^c}{{36}^{0}c-{16}^0}=-\frac{In\left(5/6\right)t}{5min}t=\frac{In\left(4/5\right)}{In\left(5/6\right)}\times 5min=6.1min$