Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the radiation pressure on an object that absorbs the light is given by
$$P=\dfrac{I}{c}\tag 1$$
where $P$ here is for pressure.
The author told us to assume that the object absorbs all the light of 1360 W/m$^2$.
The force exerted by this radiation is then given by
$$F=PA$$
where $P$ is the pressure and $A$ is the cross-sectional area of the object.
Plug from (1),
$$F=\dfrac{I}{c}A$$
We need to find the force exerted on the whole earth by this radiation, so we need to find its cross-sectional area as a sphere which is a huge circle. So $A_E= \pi R_E^2$
$$F=\dfrac{\pi R_E^2I}{c} $$
Plug the known;
$$F=\dfrac{\pi (6.37\times 10^6)^2(1360)}{(3\times 10^8)}$$
$$F=\color{red}{\bf 5.779\times 10^8}\;\rm N$$
$$\color{blue}{\bf [c]}$$
To compare this force with the gravitational force, we need to divide the radiation force over the gravitational force.
Recalling that the gravitational force is given by $F_G=Gm_Sm_E/R_{E\rightarrow S}^2$ where $R_{E\rightarrow S}$ is the distance between the Sun and the Earth.
$$\dfrac{F}{F_G}=\dfrac{(5.779\times 10^8)R^2_{E\rightarrow S}}{Gm_Em_S}$$
Plug the known;
$$\dfrac{F}{F_G}=\dfrac{(5.779\times 10^8)(150\times 10^9)^2}{(6.672\times 10^{-11})(1.99\times 10^{30})(5.98\times 10^{24})}$$
$$\dfrac{F}{F_G}=1.64\times 10^{-14}$$
Thus,
$$F=\color{red}{\bf 1.64\times 10^{-14}}F_G$$
which is too small compared to the gravitational force.
