Let the temperature of the surrounding be $\theta_{0}$.

According to Newton's law of cooling.

$\frac{62-50}{10}=\mathrm{K}\left(\frac{62+50}{2}-\theta_{0}\right)$

$\text { or } \frac{12}{10}=\mathrm{K}\left[\frac{112}{2}-\theta_{0}\right]$           $…(i)$

and $\frac{50-42}{10}=K\left[\frac{50+42}{2}-\theta_{0}\right]$

or $\frac{8}{10}=K\left[\frac{92}{2}-\theta_{0}\right]$           $…(ii)$

Divide eqn. $(i)$ by $(ii),$ we get;

$\frac{12}{10} \times \frac{10}{8}=\frac{112-2 \theta_{0}}{92-2 \theta_{0}}$ or $\frac{3}{2}=\frac{112-2 \theta_{0}}{92-2 \theta_{0}}$

or $\quad 276-6 \theta_{0}=224-4 \theta_{0}$

or  $276-224=2 \theta_{0}$ or $\theta_{0}=\frac{52}{2}=26^{\circ} \mathrm{C}$