Take into account that efficiency is given by the following expression:
![\text{ efficiency=(output work/input work)}\cdot100]()
In this case, you have:
efficieny = 28%
output work = 30J
input work = ?
Replace the previous values of the parameters into the formula for efficieny, solve for input work and simplify:
28% = (30J/input work)*100%
input work = 30J*(100% / 28%)
input work = 107.14 J
Hence,about 107.14 J was put into the machine