Answer
a) $1.812\times 10^5\;\rm W$
b) $3\times 10^8\;\rm J$
c) $\$ 773$
Work Step by Step
a)
We know that the rate of the heat transferred by conduction is given by
$$\dfrac{Q}{t}=kA\dfrac{T_1-T_2}{l}$$
And we have here 3 areas, walls, roof, and window.
So, the net heat transferred by conduction is given by
$$\left(\dfrac{Q}{t}\right)_{net}=k_{wall}A_{wall}\dfrac{T_1-T_2}{l_{wall}}+k_{roof}A_{roof}\dfrac{T_1-T_2}{l_{roof}}+k_{window}A_{window}\dfrac{T_1-T_2}{l_{window}}$$
Noting that each area has the same temperature difference since we need the inside temperature to remain at $23^\circ \rm C $ while outside the temperature was $-15^\circ \rm C $
Thus,
$$\left(\dfrac{Q}{t}\right)_{net}=\left[\dfrac{k_{wall}A_{wall}}{l_{wall}}+\dfrac{k_{roof}A_{roof}}{l_{roof}}+\dfrac{k_{window}A_{window}}{l_{window}}\right]\left(T_1-T_2\right)$$
Now we need to plug the known;
$$\left(\dfrac{Q}{t}\right)_{net}=\left[\dfrac{0.023\cdot 410}{0.195}+\dfrac{ 0.1\cdot 250}{0.055}+\dfrac{ 0.84\cdot 33}{0.65\times 10^{-2} }\right]\left(23-(-15)\right)$$
$$\left(\dfrac{Q}{t}\right)_{net}=\color{red}{\bf 1.812\times 10^5}\;\rm W$$
b) The total heat needed to warm the whole house is the heat needed to warm the air plus the heat loss due to conduction.
Thus,
$$Q_t=Q_{air}+Q_{loss}$$
$$Q_t=m_{air}c_{air}\Delta T +Q_{loss}$$
Noting that $m_{air}=\rho_{air} V_{air}$, and that $Q_{loss}=\left(\dfrac{Q}{t}\right)_{net}\times \Delta t$
$$Q_t=\rho_{air} V_{air}c_{air}\Delta T +\left(\dfrac{Q}{t}\right)_{net} \Delta t\tag 1$$
Note that the Specific Heat of air is $0.24 \;\rm kcal/kg.^\circ C$ and to conert it to Joules, we need to multiply by $\rm 4186\; J/kcal$.
In this case we assumed that the temperature difference between inside and outside is the average between the start of heating at $\rm 15^\circ C$ and the final temperature of $\rm 23^\circ C$ while the out side temperature is still $\rm -15^\circ C$.
$$\Delta T_{avg}=\dfrac{30+38}{2}=34^\circ \;\rm C$$
And hence,
$$\left(\dfrac{Q}{t}\right)_{net}=\left[\dfrac{0.023\cdot 410}{0.195}+\dfrac{ 0.1\cdot 250}{0.055}+\dfrac{ 0.84\cdot 33}{0.65\times 10^{-2} }\right]\left(34\right)$$
$$\left(\dfrac{Q}{t}\right)_{net}=\color{blue}{1.62\times 10^5}\;\rm W$$
Plugging the know into (1);
$$Q_t=\left[1.29\cdot 750\cdot 0.24\cdot 4186\cdot (23-15) \right]+\left[ 1.62\times 10^5\cdot 30\cdot 60\right]$$
$$Q_t=\color{red}{\bf3.0\times 10^8}\;\rm J$$
c)
Since one month contains 30 days, so
$$\Delta t=1\;\rm month\cdot \dfrac{30\;day}{1\;month}\cdot \dfrac{24\;h}{1\;day}\cdot \dfrac{3600\;s}{1\;h}=2.592\times 10^6\;\rm s$$
We also know that 90$\%$ of gas used in heating, so the amount of gas used is
$$ 0.9Q_{gas}=\left(\dfrac{Q}{t}\right)_{net}\cdot \Delta t $$
$$Q_{gas}=\dfrac{\left(\dfrac{Q}{t}\right)_{net}\cdot \Delta t}{0.9}$$
Plugging the know;
$$Q_{gas}=\dfrac{ 1.812\times 10^5 \cdot 2.592\times 10^6 }{0.9}=5.22\times 10^{11}\;\rm J$$
The amount of gas in kg;
$$m=5.22\times 10^{11}\;\rm J\cdot \dfrac{1\;kg}{5.4\times 10^7\;\rm J}=9666.6\;\rm kg$$
And hence, the cost per month;
$${\rm Cost}=9666.6\;\rm kg\cdot \dfrac{\$0.08}{1\;kg}=\color{red}{\$\bf773}$$
