I assume here that the engine operates following a Carnot cycle, which achieves the maximum possible efficiency.
Under this assumption, the efficiency of the engine (so, the efficiency of the Carnot cycle) is given by
[tex]\eta = 1- \frac{T_{cold}}{T_{hot}} [/tex]
where
[tex]T_{cold}[/tex] is the cold temperature
[tex]T_{hot}[/tex] is the hot temperature
For the engine in our problem, the cold temperature is 313 K while the hot temperature is 425 K, so the effiency of the engine is
[tex]\eta=1- \frac{313 K}{425 K}=0.264 = 26.4 \% [/tex]
