Answer
a) $ 3.63\times 10^6\;\rm J$
b) $ 3.05\times 10^5\;\rm J$
Work Step by Step
a) We know, for a Carnot cycle, that
$$\eta_{\rm Carnot}=1-\dfrac{T_C}{T_H}=1-\dfrac{Q_C}{Q_H}$$
Thus,
$$\dfrac{T_C}{T_H}=\dfrac{Q_C}{Q_H}$$
Hence, the heat exhausted into the room is given by
$$Q_H=\dfrac{Q_CT_H}{T_C}\tag 1$$
Now we need to find $Q_C$ which is the heat exhausted to convert the 10-kg-0$ ^\circ\rm C$ water to ice.
$$Q_C=mL_{\rm f}\tag 2$$
Plugging into (1);
$$Q_H=\dfrac{mL_{\rm f}T_H}{T_C}\tag 3 $$
Plugging the known;
$$Q_H=\dfrac{(10)(3.33\times 10^5)(25+273)}{0+273} $$
$$Q_H=\color{red}{\bf 3.63\times 10^6}\;\rm J$$
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b) The energy supplied needed to the refrigerator is given by
$$W_{in}=Q_H-Q_C$$
Plugging from (2) and (3);
$$W_{in}=\dfrac{mL_{\rm f}T_H}{T_C}-mL_{\rm f}=mL_{\rm f}\left[\dfrac{T_H}{T_C}-1\right]$$
Plugging the known;
$$W_{in} =(10)(3.33\times 10^5)\left[\dfrac{25+273}{0+273}-1\right]$$
$$W_{in} =\color{red}{\bf3.05\times 10^5}\;\rm J$$
