The answer is C) 10A Fuse.To calculate the current flowing to the kettle, you use the P=IV equation to work out that I=P/V. Putting P = 1200 and V = 230 into this equation gives I = 1200/230 = 5.217A. Therefore, any fuse labelled at less than or equal to 5.217A will blow when the kettle is operating, meaning the chosen fuse must be greater than 5.217A.Fuses are a safety device designed to prevent too much current flowing through a device. Therefore, you want the current labelled on the fuse to be above the current expected to flow through the device, but not too much above. In this case, 13A is too far above the expected current of 5.217A to be an appropriate choice. This leaves you with only one option - the 10A fuse - which is above the expected current, but still close enough to it to function as a safety device, making it the most appropriate choice.