Correct Answer - `80.9%`
Here`V= 110 V, l = 2A`
power supplied to motor `= 1//1 = 110 xx 2 = 220 W`
power loss in the form of heat `= 10 cals^(-1)`
`= 10 xx 4.2 Js^(-1) = 42 Js^(-1)`
power used by motor `= 220 - 42 = 178 W`
Effeciency `eta = ("power used by motor")/("power supplied to motor") xx 100`
`= (178)/(220) xx 100 = 80.9%`