1. Magnetic field inside the long solenoid

For a long solenoid, the magnetic field inside is uniform and along its axis:

$B = \mu_0 n I$

The electron is fired along the radius of the solenoid, while the magnetic field is along the axis. Hence the velocity is perpendicular to the field.

2. Motion of electron in magnetic field

When a charged particle enters a magnetic field with velocity perpendicular to $\vec B$, it moves in a circle of radius

$r = \frac{mv}{eB}$

where $e$ is the magnitude of charge of electron.

So here,

$r = \frac{mv}{e\mu_0 n I}$

3. Condition for not hitting the inner surface of the solenoid

The electron starts from the axis of the solenoid and moves initially along the radius. Due to magnetic force, it follows a circular path in the plane perpendicular to the magnetic field.

Since it starts from the center, the circular trajectory is tangent to the initial radial direction at the center. Therefore the center of the circular path lies at a distance $r$ from the axis, and the farthest distance reached from the solenoid axis is

$2r$

For the electron to not hit the cylindrical surface of radius $R$,

$2r \le R$

At the maximum possible speed,

$2r = R$

Thus,

$2\left(\frac{mv}{e\mu_0 n I}\right)=R$

$v = \frac{e\mu_0 n I R}{2m}$

4. Matching with the options

This corresponds to:

$\boxed{\frac{e\mu_0 n I R}{2m}}$

So the correct option is C.