Answer
$5.34\times 10^4\;\rm J$
Work Step by Step
We know that the heat energy exhausted $Q_H$ into the room by the water is given by
$$W_{in}=Q_H-Q_C$$
Hence,
$$Q_H=W_{in}+Q_C\tag 1$$
And we also know that the work is given by
$$K=\dfrac{Q_C}{W_{in}}$$
where $K$ is the performance coefficient of a refrigerator.
Hence,
$$W_{in}=\dfrac{Q_C}{K}$$
Plugging into (1);
$$Q_H=\dfrac{Q_C}{K}+Q_C$$
$$Q_H=Q_C\left[\dfrac{1}{K}+1\right]$$
where $K=4$, so
$$Q_H=\frac{5}{4}Q_C\tag 2$$
Now we need to find $Q_C$ which is given in three stages since the water losses heat in these three stages:
1- cooling down from 15$^\circ$C to 0$^\circ$C.
$$Q_{C1}=mc_{water}\Delta T_1$$
2- converting to a piece of ice at 0$^\circ$C.
$$Q_{C2}=mL_{\rm f}$$
3- cooling down from 0$^\circ$C to -15$^\circ$C.
$$Q_{C3}=mc_{ice}\Delta T_2$$
Thus,
$$Q_C=Q_{C1}+Q_{C2}+Q_{C3}$$
$$Q_C=mc_{water}\Delta T_1+mL_{\rm f}+mc_{ice}\Delta T_2 $$
$$Q_C=m\left[c_{water}\Delta T_1+ L_{\rm f}+ c_{ice}\Delta T_2\right]\tag 3$$
Now we need to find the mass of the water which is given by
$$\rho=\dfrac{m}{V}\;\;\;\;\rightarrow\;\;\;\;m=\rho V$$
Plugging into (3);
$$Q_C=\rho V\left[c_{water}\Delta T_1+ L_{\rm f}+ c_{ice}\Delta T_2\right]$$
Plugging into (2);
$$Q_H=\frac{5}{4}\rho V\left[c_{water}\Delta T_1+ L_{\rm f}+ c_{ice}\Delta T_2\right]$$
Plugging the known;
$$Q_H=\frac{5}{4}(1000)(100\times 10^{-6})\left[(4186)(15)+(3.33\times 10^5)+ (2090)(15)\right]$$
$$Q_H=\color{red}{\bf 5.34\times 10^4}\;\rm J$$
