Answer:
The steady-state temperature is
Yes it is chilly
Step-by-step explanation:
From the question we are told
The intensity of solar radiation is

Generally the Stefan Boltzmann Law is mathematically represented as

Where
is the total power radiated
is the surface area of the object
is the emissivity
T is the temperature of the object
is the Boltzmann constant with a value

Generally at steady state the input power to the object is equal to the output power from the object
i.e

Now
which is the input power to the object is not dependent on the object temperature and on the Boltzmann constant
thus
is mathematically represented as

Where
is absorptive surface area mathematically represented as

Thus

And
which is the output power to the object is mathematically represented a

Where
is the radiative surface area which is mathematically as

So

=>

=>
![T = \sqrt[4]{(I)/(4 \sigma ) }](https://img.qammunity.org/2021/formulas/physics/college/mo6368of4am0v11u0i0391pz725b4emh2f.png)
substituting values
![T = \sqrt[4]{(1370)/(4 * 5.670 *10^(-8) ) }](https://img.qammunity.org/2021/formulas/physics/college/8i4v2uqyf3547ynt1eacebv455r5dqq73c.png)

Converting to degrees


This implies that at steady state it is chilly