The correct answer is $\boxed{\text{B}}$.
The horizontal range of a projectile is given by the following formula:
$$R = \frac{v^2 \sin^2 \theta}{g}$$
where $v$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
In this case, $v = 98 \text{ m/s}$ and $\theta = 45^\circ$. Plugging these values into the formula, we get:
$$R = \frac{(98 \text{ m/s})^2 \sin^2 (45^\circ)}{9.8 \text{ m/s}^2} = 490 \text{ m}$$
Option A is incorrect because it is the initial velocity of the projectile. Option C is incorrect because it is the square of the initial velocity of the projectile. Option D is incorrect because it is the product of the initial velocity of the projectile and the sine of the angle of projection.