According to $Newton's$ law of cooling,

$\frac{{dT}}{{dt}} = K\left( {T – {T_s}} \right)$

For two cases,

$\frac{{d{T_1}}}{{dt}} = K\left( {{T_1} – {T_s}} \right)\,and\,\frac{{d{T_2}}}{{dt}} = K\left( {{T_2} – {T_s}} \right)$

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

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

${T_2} = \frac{{2T + T'}}{{2}}and\frac{{d{T_2}}}{{dt}} = \frac{{2T – T'}}{{10}}$

$So,\,\frac{T}{{10}} = K\left( {2.5\,T – T} \right)$                     $…(i)$

$\frac{{2T – T'}}{{10}} = K\left( {\frac{{2T + T'}}{2} – T} \right)$                $…(ii)$

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

$\frac{T}{{2T – T'}} = \frac{{\left( {2.5T – T} \right)}}{{\left( {\frac{{2T + T'}}{2} – T} \right)}}$

$\frac{{2T + T'}}{2} – T = \left( {2T – T'} \right) \times \frac{3}{2}$

$T' = 3\left( {2T – T'} \right)\,\,or,\,\,4T' = 6T\,\,\therefore \,\,T' = \frac{3}{2}T$