According to Newton’s law of cooling,

${{dT} \over {dt}} = K\left( {T - {T_s}} \right)$

For two cases,

${{d{T_1}} \over {dt}} = K\left( {{T_1} - {T_s}} \right)$

and ${{d{T_2}} \over {dt}} = K\left( {{T_2} - {T_s}} \right)$

Here, ${T_s} = T,{T_1} = {{3T + 2T} \over 2} = 2.5T$

and ${{d{T_1}} \over {dt}} = {{3T - 2T} \over {10}} = {T \over {10}}$

${T_2} = {{2T + T'} \over 2}$ and ${{d{T_2}} \over {dt}} = {{2T - T'} \over {10}}$

So ${T \over {10}} = K\left( {2.5T - T} \right)$   ...(i)

${{2T - T'} \over {10}} = K\left( {{{2T + T'} \over 2} - T} \right)$   ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

${T \over {2T - T'}} = {{\left( {2.5T - T} \right)} \over {\left( {{{2T + T'} \over 2} - T} \right)}}$

${{2T + T'} \over 2} - T = \left( {2T - T'} \right) \times {3 \over 2}$

$T' = 3\left( {2T - T'} \right) \Rightarrow 4T' = 6T$

$\therefore T' = {3 \over 2}T$