a
Let $T_s$ be the temperature of the surroundings.
According to $Newton's$ law of cooling
$\frac{{{T_1} - {T_2}}}{t} = K\left( {\frac{{{T_1} + {T_2}}}{2} - {T_s}} \right)$
For first $5\,minutes,$
${T_1} = {70^ \circ }C,{T_2} = {60^ \circ }C,t = 5\,minutes$
$\therefore \frac{{70 - 60}}{5} = K\left( {\frac{{70 + 60}}{2} - {T_s}} \right)$
$\frac{{10}}{5} = K\left( {65 - {T_s}} \right)$ $...(i)$
For next $5\,minutes$
${T_1} = {60^ \circ }C,{T_2} = {54^ \circ }C,t = 5 minutes$
$\therefore \frac{{60 - 54}}{5} = K\left( {\frac{{60 + 54}}{2} - {T_s}} \right)$
$\frac{6}{5} = K\left( {57 - {T_s}} \right)$ $...(ii)$
Divide eqn. $(i)$ by eqn. $(ii)$ , we get
$\frac{5}{3} = \frac{{65 - {T_s}}}{{57 - {T_s}}}$
$285 - 5{T_s} = 195 - 3{T_s}$
$2{T_s} = 90\,\,or\,\,{T_s} = {45^ \circ }C$