Look for the gamma rays
Let's start off with a couple equations:
The Schwarzschild radius of a black hole:
$r_s = {2 G M\over c^2}$
The power radiated via Hawking radiation:
$P = \frac{\hbar c^6}{15360\pi G^2 M^2} = \frac{\hbar c^2}{3840\pi r_s^2}$
And we'll make use of the evaporation time too:
$t_{ev} = \frac{5120\pi G^2 M^3}{\hbar c^4} = \frac{640 \pi c^2 r_s^3}{\hbar G}$
So if we plug in 2.5 attometer for $r_s$, we can find the power emitted by your black hole. It comes out to be around 1.257 * 10^14 Watts or 125.7 TW. Also plugging into the evaporation time we can find that this black hole will have a lifetime of around 12700 years, so there's plenty of time to spot this thing (especially since it'll be continually increasing in brightness unless something/someone is feeding it).
This ends up being equivalent to a black-body with a temperature of around 7.3 * 10^13 K. So this thing is predominantly radiating gamma radiation.
But how bright is this really? For that we need to work out its apparent magnitude. We can do this by computing its intensity compared to the sun. I'll assume the black hole is about 10 billion km away (for reference Pluto is about 7.5 billion km from Earth on average). This is a bit of an abuse of apparent magnitude since it usually refers to visible light and I'm using the entire radiated power of the black hole.
Using that as a reference, I get an apparent magnitude of about 13 (the sun's apparent magnitude is -27 and this black hole is about 40 steps above that, so 13).
That makes this thing about 2.5 times brighter than Pluto or about as bright as this quasar, so you definitely won't be seeing it with the naked eye. However, it should be pretty noticeable if you point a gamma ray detector at it I would think.