The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by:

$R = \frac{v^2}{g} \sin(2\theta)$

where $g$ is the acceleration due to gravity.

From this equation, we can see that the range is dependent on the sine of twice the launch angle.

Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 \times 15^\circ)$ and $\sin(2 \times 45^\circ)$:

$\sin(30^\circ) = \frac{1}{2}$

$\sin(90^\circ) = 1$

Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$, because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$.

So, the range when the projectile is launched at $45^\circ$ will be $2 \times 50$ m = $100$ m.

So, $100$ m is the correct answer.