What is the escape speed for an electron initially at rest?

I'm assuming that by 'uniformly distributed' you're referring to the volume of the sphere, not its surface. (In mathematics, "sphere" means the 2-dimensional surface of a "ball," which is 3-dimensional; but this being physics, I will take that as a volume distribution.) Although, for this geometry, the potential outside the sphere will be the same either way.

To find the escape speed, v, you just need the electrical potential, V, at that surface. Mathematically, this is just like finding the gravitational potential (= potential energy per mass) at the surface of a spherical mass. Instead of G for the gravitational case, there will be 1/4πε₀. So instead of V(r) = GM/r, you'll have:

V(r) = Q/(4πε₀r) = (1.5•10ˉ¹⁵C / 1.0•10ˉ²m)•8.99•10⁹ N•m²/C²

= 1.35•10ˉ³V [*]

The potential energy of the electron is

PE(r) = qV(r) = -eV(r) = -1.602•10ˉ¹⁹C • 1.35•10ˉ³V = -2.16•10ˉ²²J

So the electron needs a kinetic energy equal to (-PE) to escape.

KE = ½mv² = 2.16•10ˉ²²J

v = √(2KE/m) = √(4.32•10ˉ²²/0.911•10ˉ³⁰) m/s

= 2.18•10⁴m/s ≈ 22 km/s

[*] Some units manipulations:

C•V = J = N•m

... so ...

(C/m)•N•m²/C² = N•m/C = J/C = V