We are given that power of the electric iron, `P = 1200 W =1.2 kW`,
total time for which the electric iron is used in the month of April `(30 days)` =` (30 min//day) (30day)`
`=((1)/(2)h//day) (30day) = 15 h`
Energy consumed, `E = Pt = (1.2 kW) (15h) = 18kWh (as P = "work done//time" = E//t)`
