#### Answer

61%; this is much larger than the 25 percent actual efficiency.

#### Work Step by Step

As we know that
$T_C=90F^{\circ}=(90-32)\frac{5}{9}+273K=305K$
$e_{Carnot}=(1-\frac{T_C}{T_K})\times 100\%$
We plug in the known values to obtain:
$e_{Carnot}=(1-\frac{305K}{783K})\times 100\%$
$e_{Carnot}=61\%$
It is clear that the efficiency is greater than actual efficiency that is $25\%$.