A reversible engine and a reversible heat pump running between the same two reservoirs share the exact same temperature ratio, because reversibility fixes $Q_L/Q_H = T_L/T_H$ regardless of which direction the cycle runs. This lets us reuse the efficiency information from the engine mode directly in the heat pump mode.
As an engine, $\eta = 1 - T_L/T_H = 0.55$, so $T_L/T_H = 0.45$. This is the same ratio that connects the heat pump's cold-side heat intake $Q_L$ and hot-side heat rejection $Q_H$, since $Q_L/Q_H = T_L/T_H$.
Now with the engine reversed into a heat pump, the heat absorbed from the low temperature reservoir is given as $Q_L = 45$ kJ/cycle. Rearranging the ratio gives $Q_H = Q_L \times (T_H/T_L) = 45 / 0.45$.
This works out to exactly $100$ kJ/cycle, so the heat pump rejects $100$ kJ/cycle to the high temperature reservoir, which is the required answer.