Answer
$134.6^oC$
Work Step by Step
We have $T_H =400^oC =673K$ and $\eta = 40\% =0.4$
$\eta = 1- T_{Ci}/T_H$
$0.4 = 1- T_{Ci}/673K$
$ T_{Ci} =403.8K = 130.8^oC$
Now when $\eta =60\% = 0.6$
$\eta = 1- T_{Cf}/T_H$
$0.6 = 1- T_{Cf}/673K$
$T_{Cf} = 269.2K =-3.8^oC$
Hence $T_{Cf} -T_{Ci} = 173.8^oC - (-3.8^oC) =134.6^oC$
Thus temperature of cold reservoir should be decreased by $134.6^oC$ to increase efficiency to $60\%$
