The same amount of heat comes out from the heat source and is absorbed by the lower-temperature body.
Now the rate of change in entropy during the heat coming from the source is \(\Delta {S_1} = - \frac{Q}{{{T_1}}}\).
The rate of change in entropy during the absorption of heat is \(\Delta {S_2} = \frac{Q}{{{T_2}}}\).
Therefore the rate of change of net entropy is,
\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = - \frac{Q}{{{T_1}}} + \frac{Q}{{{T_2}}}\\ = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\end{array}\)
Now substituting the values in the above equation, you get,
\(\begin{array}{c}\Delta S = Q\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)\\ = \left[ {\left( {8.40 \times 4.186} \right)\;{\rm{J/s}}} \right] \times \left( {\frac{1}{{295\;{\rm{K}}}} - \frac{1}{{498\;{\rm{K}}}}} \right)\\ = 4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\end{array}\)
Hence, the rate of increase of entropy is \(4.86 \times {10{ - 2}}\;\frac{{{\rm{J/K}}}}{{\rm{s}}}\).