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The initial temperature is \({T_1} = 210^\circ {\rm{C}}\).

The final temperature is \({T_2} = 45^\circ {\rm{C}}\).

The power output is \(P = 910\;{\rm{W}}\).

In this problem, while calculating the rate of heat input, the efficiency of Carnot engine will be utilized because the operating engine is perfect Carnot.

The relation of efficiency is given by,

\({\eta _{\rm{C}}} = 1 - \frac{{{T_2}}}{{{T_{\rm{1}}}}}\)

On plugging the values in the above relation.

\(\begin{aligned}{c}{\eta _{\rm{C}}} &= 1 - \left( {\frac{{\left( {45^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( {210^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\{\eta _{\rm{C}}} &= 0.34\end{aligned}\)

The relation of efficiency is given by,

\(\begin{aligned}{l}{\eta _{\rm{C}}} &= \frac{W}{{W + Q}}\\Q &= W\left( {\frac{1}{{{\eta _{\rm{C}}}}} - 1} \right)\\\frac{Q}{t} &= \frac{W}{t}\left( {\frac{1}{{{\eta _{\rm{C}}}}} - 1} \right)\\\frac{Q}{t} &= P\left( {\frac{1}{{{\eta _{\rm{C}}}}} - 1} \right)\end{aligned}\)

Here, \(\frac{Q}{t}\) is the rate of heat output and W is the work done.

On plugging the values in the above relation.

\(\begin{aligned}{l}\frac{Q}{t} &= \left( {910\;{\rm{W}}} \right)\left( {\frac{1}{{0.34}} - 1} \right)\\\frac{Q}{t} &= 1766.4\;{\rm{W}}\end{aligned}\)

Thus, \(\frac{Q}{t} = 1766.4\;{\rm{W}}\) is the required rate of heat output.