Answer
a) $1.73\times10^{17}\;\rm W$
b) $4.74^\circ \;\rm C$
Work Step by Step
a)
To find the rate at which the whole Earth receives energy from the Sun, we need to recall that
$$\dfrac{Q}{t}=A\cdot \text{Solar constant}$$
The area of Earth that receives heat from the sun is the cross-sectional area of Earth which is a huge circle with a radius of $R_E$.
$$\dfrac{Q}{t}=\pi R_E^2\cdot \text{Solar constant}=\pi \cdot \left(6.38\times10^6\right)^2\cdot 1350$$
$$\dfrac{Q}{t}=\color{red}{\bf1.73\times10^{17}}\;\rm W\tag 1$$
b)
Now we need to find the temperature of Earth when it is emitting the same amount of energy back to space.
$$\dfrac{Q}{t}=\sigma\epsilon A\left[T_E^4-T_{space}^4\right]$$
Solving for $T_E$;
$$\dfrac{Q}{t}\cdot\dfrac{1}{\sigma\epsilon A}= T_E^4-T_{space}^4 $$
$$T_E^4=\dfrac{Q}{t}\cdot\dfrac{1}{\sigma\epsilon A}+T_{space}^4 $$
$$T_E =\sqrt[4]{\dfrac{Q}{t}\cdot\dfrac{1}{\sigma\epsilon A}+T_{space}^4 }$$
Now since the whole Earth emits heat, the area here is the area of a sphere $A=4\pi r_E^2$
$$T_E =\sqrt[4]{\dfrac{Q}{t}\cdot\dfrac{1}{\sigma\epsilon \cdot 4\pi r_E^2}+T_{space}^4 }$$
Plugging the know and from (1);
$$T_E =
\sqrt[4]{\left[1.73\times10^{17} \cdot\dfrac{1}{5.67\times10^{-8}\cdot 1\cdot 4\pi\cdot \left(6.38\times10^6\right)^2}\right]+2.7^4 }$$
$$T_E=277.9\;\rm K\approx \color{red}{\bf4.74}^\circ\rm C$$
