Impulse can be written as either of two popular expressions:
$I=F \Delta t=m \Delta v$

From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:
$E=U_i+K_i=U_f+K_f$
Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:
$m g h_i=\frac{1}{2} m v_f^2$
Canceling out mass and rearranging for final velocity, we get:
$v_f=\sqrt{2 g h_i}$
We know these variables, allowing us to solve for the velocity:
$v_f=\sqrt{2\left(10 \frac{m}{s^2}\right)(1.2 m)}=4.9 \frac{\mathrm{m}}{\mathrm{s}}$
Plugging this value into the expression for impulse, we get:
$I=(0.1 \mathrm{~kg})\left(4.9 \frac{\mathrm{m}}{\mathrm{s}}\right)=0.49 \mathrm{~N} \cdot \mathrm{s}$