Solution:

The electric field \(\vec{E}\) is related to the potential \(V\) by:
\[
\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}\right).
\]

Given \(V = k[2x^2 + y^2 - 2z]\):

1. Partial derivatives:
- \(\frac{\partial V}{\partial x} = 4kx\),
- \(\frac{\partial V}{\partial y} = 2ky\),
- \(\frac{\partial V}{\partial z} = -2k\).

2. Electric field components at \((1, 0, 1)\):
- \(E_x = -\frac{\partial V}{\partial x} = -4k(1) = -4k\),
- \(E_y = -\frac{\partial V}{\partial y} = -2k(0) = 0\),
- \(E_z = -\frac{\partial V}{\partial z} = -(-2k) = 2k\).

3. Magnitude of \(\vec{E}\):
\[
|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-4k)^2 + (0)^2 + (2k)^2} = \sqrt{16k^2 + 4k^2} = \sqrt{20k^2} = 2k\sqrt{5}.
\]

Thus, the magnitude of the electric field is \(2k\sqrt{5}\).