Answer
a) $5.05\times 10^4\;\rm J$
b) $32\;\rm min$
Work Step by Step
a)
First of all, we need to find the removed heat from the water to cool it from 25$^\circ$C to -17$^\circ$C.
$$Q_L=\overbrace{ m_{water}c_{water}\Delta T_1}^{{\color{red}\;\text{cooling water to 0 C}}}+\overbrace{ mL_f}^{{\color{red}\;\text{converting water into ice}}}+\overbrace{ m_{ice}c_{ice}\Delta T_2 }^{{\color{red}\;\text{Cooling ice to -17 C}}}$$
whereas $\Delta T_1$ is the temperature difference from the room temperature of 25$^\circ$C to the freezing point 0$^\circ$C, $\Delta T_2$ is the temperature difference from the freezing point of 0$^\circ$C to -17$^\circ$C, $L_f$ is the latent heat of fusion required to convert the whole amount of water into ice.
Now we need to plug the given;
$$Q_L=\left[0.65\cdot 4186\cdot 25\right]+[0.65\cdot 333\times 10^3]+\left[0.65\cdot 2100\cdot 17\right]$$
$$Q_L=3.077\times10^5\;\rm J\tag 1$$
Now we need to find the work by using the Carnot efficiency
$$e=\dfrac{T_H-T_L}{T_H}=\dfrac{W}{Q_H}$$
whereas $Q_H=Q_L+W$;
$$ \dfrac{T_H-T_L}{T_H}=\dfrac{W}{Q_L+W}$$
Solving for $W$;
$$\left( \dfrac{T_H-T_L}{T_H}\right)\left(Q_L+W\right)= W$$
$$ Q_L\dfrac{T_H-T_L}{T_H}+W\dfrac{T_H-T_L}{T_H} = W$$
$$W-W\dfrac{T_H-T_L}{T_H}=Q_L\dfrac{T_H-T_L}{T_H}$$
$$W\left(1-\dfrac{T_H-T_L}{T_H}\right)=Q_L\dfrac{T_H-T_L}{T_H}$$
$$W\left(1-\left[1-\dfrac{ T_L}{T_H}\right]\right)=Q_L\dfrac{T_H-T_L}{T_H}$$
$$W \dfrac{ T_L}{T_H} =Q_L\dfrac{T_H-T_L}{T_H}$$
$$W =Q_L\dfrac{T_H-T_L}{T_H}\cdot \dfrac{T_H}{ T_L}$$
$$W =Q_L\dfrac{T_H-T_L}{ T_L}$$
Plugging $Q_L$ from (1) and plugging the known; (Remember to convert the temperatures into Kelvin).
$$W =3.077\times10^5\cdot \dfrac{298-256}{ 256}$$
$$W=\color{red}{\bf 5.05\times 10^4}\;\rm J$$
b)
We know that the power is given by
$$P=\dfrac{W}{\Delta t}$$
Solving for $\Delta t$ to find the time interval the compressor takes to complete the process.
$$\Delta t=\dfrac{W}{P}=\dfrac{5.05\times 10^4}{105}=481\;\rm s$$
and since the compressor works only 25$\%$ of the time, so the whole time it takes is 4 times if he worked without a stop.
So that;
$$\Delta t_{tot}=\dfrac{481}{0.25}=1924\;\rm s\approx \color{red}{\bf32}\;\rm min$$