White light Hawking radiation If the Hawking radiation is similar in spectrum to a black body, then at what mass of the black hole its radiation would have the same peak energy as the sunlight?
 A: The equivalent temperature of a black hole (as seen from infinity- since the blackbody radiation will be red-shifted as it moves away from the black hole) is given by $$\frac{1}{8\pi M}$$ in natural units. Or with all the constants in there: $$\frac{\hbar c^3}{8\pi k_BGM}$$
For the Sun's temperature of $5778\rm K$, this corresponds to $2\cdot 10^{19} \rm kg$. Or about 3 millionths the mass of the earth. Impossibly small for a black hole, of course.
Edit: requested in comment
The Schwarzschild radius for the black hole would be $31\rm nm$. Since it's blackbody radiation, the power should be determined by the Stefan-Boltzmann law, so the luminosity is given by: $$ L = 4\pi r^2 \sigma T^4$$
Plugging this in gives a total luminosity of around 800 nanowatts. So very tiny. To find out the lifetime, we can set up an differential equation: $$\frac{dM}{dt}=-\frac{L}{c^2}=-\frac{4\pi r^2\sigma T^4}{c^2}$$
Changing to natural units so this doesn't get super long (note that $\sigma=\frac{\pi^2}{60}$ in natural units) and substituting in $r$ and $T$: $$\frac{dM}{dt}=-\frac{\pi^3(2M)^2\left(\frac{1}{8\pi M}\right)^4}{15}=-\frac{1}{15360\pi M^2}$$
Rearranging this and integrating: $$\int_{M_0}^0M^2dM=-\int_0^t\frac{dt'}{15360\pi}$$
$$\frac{M^3}{3}=\frac{t}{15360}$$ $$t=5120M^3$$
Putting the constants back in: $$t=\frac{5120G^2M^3}{\hbar c^4}$$
For our initial numbers, this gives you $8\cdot 10^{33} \rm yr$. So it comes out to be an extremely dim light bulb that will be around for an unimaginably long time.
For fun: a black hole the mass of the Empire State building would last for thirty years and would start out with a luminosity of 3.2 petawatts, about 500 times the world's power usage, or 100 small nuclear bombs per second. And it would get brighter over the thirty year period.
