The first step in solving this problem is to calculate the change in momentum of the ball when it is caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in velocity (as momentum is mass times velocity).

The ball is initially moving with a velocity of $v_i = 25 \mathrm{~m/s}$ before the catch and finally comes to rest with a velocity of $v_f = 0 \mathrm{~m/s}$ after the catch. Since the ball is caught, the final velocity is zero. The change in velocity $$\Delta v = v_f - v_i = 0 - 25 = -25 \mathrm{~m/s}.$$ Remember that the direction of the force exerted by the ball on the hand will be opposite to the direction of the ball's initial motion.

The mass of the ball $m$ is given as $120 \mathrm{~g}$ which needs to be converted into kilograms to maintain SI units: $$m = 120 \mathrm{~g} = 120 \times 10^{-3} \mathrm{~kg} = 0.12 \mathrm{~kg}.$$

Now we can calculate the change in momentum (impulse): $$\Delta p = m \Delta v = 0.12 \mathrm{~kg} \times (-25 \mathrm{~m/s}).$$

Substituting the values we get: $$\Delta p = 0.12 \times -25 = -3 \mathrm{~kg \cdot m/s}.$$

The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial motion, which makes sense because the ball's velocity is reduced to zero.

The magnitude of the impulse is independent of the sign and is $3 \mathrm{~kg \cdot m/s}$.

Impulse is also equal to the average force exerted on the ball times the time interval during which the force is exerted. We can use the formula: $$\Delta p = F_{avg} \Delta t$$

Where $F_{avg}$ is the average force and $\Delta t$ is the time interval of $0.1 \mathrm{~s}$. Re-arranging the formula to solve for $F_{avg}$ gives us: $$F_{avg} = \frac{\Delta p}{\Delta t}.$$

Substituting the known values we have: $$F_{avg} = \frac{3}{0.1} = 30 \mathrm{~N}.$$

The magnitude of the average force exerted by the hand of the player to catch the ball is $30 \mathrm{~N}$.