The Apparent size of some celestial object viewed from another celestial object is also known as angular diameter . The formula for angular diameter is: $δ=206265∗(actualDiameter/distance)$
Also generally it is in terms of arcseconds. 206265 is the conversion factor for arcsec however if you want the answer to be in degrees then the conversion factor would be 57.3. The actual diameter would be 12742.018 kilometers (equatorial) and the distance would be 1 AU/149597871 kilometers therefore the Earth's angular diameter would be 17.866808 arcsec / 0.00496336 degrees.
However since the Earth travels in an elliptical orbit due to Kepler's law of planetary motion the distance and therefore apparent size may differ depending on the season. The 1 AU distance is the Semi-Major axis of Earth's orbit around the Sun
One can use the Solar constant for the power of the light and multiply it with the cross sectional area/2d area of the earth.
Also the Earth both absorbs and reflects the light so even in that tiny fraction of the light emitted by the Sun so even in that tiny fraction only a few encounter the Earths surface the rest is absorbed by Green(hous) gases releasing Thermal energy/Heat.
So the cross-sectional area of earth is $1.275165×10^8$ square kilometers which when multiplied by the Solar constant gives $174$ petawatts!
To compare according to a 2010 estimate the world uses $18.47$ trillion kilowatt hours per year! but when you divide it by $174$ petawatts it gives $82525$ (years)!