Answer
$-18^\circ \rm C$
Work Step by Step
To calculate the earth's temperature without an atmosphere, we can use the Stefan-Boltzmann law, which relates the temperature of an object to the amount of radiation it emits. The law states that the power radiated by a blackbody is proportional to the fourth power of its absolute temperature (in Kelvin):
$$P =e \sigma A T^4\tag 1$$
where $P$ is the power radiated, $A$ is the surface area of the object, $e$ is its emissivity (which is given as 1), $T$ is its temperature in Kelvin, and $\sigma$ is the Stefan-Boltzmann constant, which is equal to $5.67 \times 10^{-8} \rm \; W/m^2K^4$
We know that the intinisty is given by
$$I=\dfrac{P}{A}$$
where $P$ here is the power NOT the pressure.
The power absorbed by the earth is given by:
$$P_{absorbed} = 0.7IA_1 \tag 2$$
where $A_1$ is the area of half of the earth's surface since this is the par of the death that faces sun's radiation.
In Equlibriium the absorbed power is equal to the radiated power. So, $P=P_{absorbed}$
Thus,
$$e\sigma A T^4= 0.7IA_1 $$
Hence,
$$T=\left[\dfrac{0.7IA_1}{e\sigma A }\right]^{\frac{1}{4}}$$
where $A$ is the area of the full earth's surface which is given by $4\pi R_E^2$ since it radiates from the whole surface, while $A_1=\pi R_E^2$ since it absorbs the energy from one half of its surface (noting that the area of half the Earth's surface is $2\pi R_E^2$ but $A_1$ is the effective area, we can reach that as imagine that the power given to the Earth will be equal to the intensity times the area of the shadow created by the Earth. It is obvious that this shadow is a sphere of radius $R_E$).
$$T=\left[\dfrac{0.7I(\color{red}{\bf\not} \pi \color{red}{\bf\not} R_E^2)}{e\sigma (4\color{red}{\bf\not} \pi \color{red}{\bf\not} R_E^2) }\right]^{\frac{1}{4}}$$
$$T=\left[\dfrac{0.7 I}{4e\sigma }\right]^{\frac{1}{4}}$$
Plugging the known;
$$T=\left[\dfrac{0.7 (1370)}{4(1)(5.67 \times 10^{-8}) }\right]^{\frac{1}{4}}=\bf 255\;\rm K$$
$$T=\color{red}{\bf -18}^\circ \rm C$$
