Intermediate fragment size can be calculated from the given MTU. Data going through the router is the payload(actual data) + headers. Since the fragment size is required to be a multiple of 8 ,the actual size of data would be $(MTU - size(header)) - ((MTU-size(header))%8)$.
No. of fragments can be calculated by $\frac{size(packet)-size(headers))}{size(fragment)}$, round it to the next integer value. So in this case a total of 5 packets. Each packet except the last one will be transmitted using the complete MTU size, so the last fragment size would be $(size(packet)-size(headers)) - ((no. of fragments-1)*size(fragment))$, which is 156 in this case. So $X+Y=456+156=612$