From Newton’s law of cooling

$\frac{dT}{dt}=k\left(T_{avg}-T_0\right)$, here, $T_0$ is surrounding temperature.

For case $1$:

$\frac{60-40}{6}=k\left(\frac{60+40}{2}-10\right)$

$\frac{20}{6}=k\left(50-10\right) ...\left(1\right)$ 

For case $2$:

$\frac{40-T}{6}=k\left(\frac{40+T}{2}-10\right) ...\left(2\right)$ 

From equation $\left(1\right)$, we have $k=\frac{1}{12}$

Putting in equation $\left(2\right)$

$\frac{40-T}{6}=\frac{1}{12}\left(\frac{40+T}{2}-10\right)$

$$\begin{gathered} \Rightarrow 10+\frac{T}{2}=80-2T \\ \Rightarrow 2T+\frac{T}{2}=70 \end{gathered}$$

$\Rightarrow T=\frac{140}{5}=28°\mathrm{C}$