$or,\,\,\frac{{{T_2}}}{{{T_1}}} = 1 - \eta  = 1 - \frac{{40}}{{100}} = \frac{3}{5}$

$\therefore {T_1} = \frac{5}{3} \times {T_2} = \frac{5}{3} \times 300 = 500\,K.$

$Increases\,in\,efficiency = 50\% \,of\,40\%  = 20\% $

$New\,efficiency,\eta ' = 40\%  + 20\%  = 60\% $

$\therefore \,\,\frac{{{T_2}}}{{{T_1}'}} = 1 - \frac{{60}}{{100}} = \frac{2}{5}$

${T_1}' = \frac{5}{2} \times {T_2} = \frac{5}{2} \times 300 = 750\,K.$

$Increases\,in\,temperature\,of\,source = {T_1}' - {T_1}$

$ = 750 - 500 = 250\,K.$