b
Assuming the temperature of surrounding be $t^{o} C$
$CASE-1$
$\frac{(60-50)}{4 \min }=k(55-t)$ $(1)$
Again, $CASE- 2$
$\frac{(40-30)}{8 \min }=k(35-t)$ $(2)$
$\frac{(1)}{(2)}=\frac{\frac{10}{4}}{\frac{10}{8}}=\frac{(55-t)}{(35-t)}$
$\Rightarrow 2(35-t)=55-t$
$\Rightarrow 70-2 t=55-t$
$\Rightarrow t=15^{\circ} \mathrm{C}$