Answer
60 K.
Work Step by Step
Find the original high temperature from the information given. The original Carnot efficiency was 25 percent.
$$0.25=1-\frac{T_L}{T_{H1}}=1-\frac{293K}{T_{H1}}$$
$$T_{H1}=390.67K$$
Find the new high temperature from the desired Carnot efficiency of 35 percent.
$$0.35=1-\frac{T_L}{T_{H1}}=1-\frac{293K}{T_{H2}}$$
$$T_{H2}= 450.77 K$$
The difference is 60 kelvin.