What is the magnitude of the component of the electric field...

By definition, the electric potential $V$ is constant everywhere on an equipotential surface.

The component of the electric field along any direction is given by $E = -\dfrac{dV}{dr}$.

For any displacement $dr$ parallel to the equipotential surface, the change in potential $dV = 0$.

Therefore, the component of the electric field parallel to the equipotential surface is $E_{\parallel} = -\dfrac{0}{dr} = 0$.

This implies that the electric field is always perpendicular to the equipotential surface, and its parallel component is zero.

Answer: Zero