1. Given data

• Mass of particle:
  $m=20\text{ g}=0.02\text{ kg}$
• Initial speed at $A$:
  $u=5\text{ m/s}$
• Surface is frictionless
• We need angular momentum of the particle about point $O$ when it reaches $B$.

2. Speed at point $B$ using energy conservation

Since the surface is frictionless, mechanical energy is conserved.

If point $A$ is at height $h$ above $B$, then $\frac12 mu^2 + mgh = \frac12 mv_B^2$

So, $v_B^2 = u^2 + 2gh$

From the figure, the vertical drop from $A$ to $B$ is $h=\frac{15}{4}\text{ m}$

Hence, $v_B^2 = 5^2 + 2(10)\left(\frac{15}{4}\right)$ $v_B^2 = 25 + 75 = 100$ $v_B = 10\text{ m/s}$

3. Angular momentum about $O$ at point $B$

Angular momentum magnitude about $O$ is $L = m v_B \times d$ where $d$ is the perpendicular distance from point $O$ to the line of motion at $B$.

From the figure, this perpendicular distance is $d=30\text{ m}$

Therefore, $L = 0.02 \times 10 \times 30$ $L = 6\text{ kg-m}^2\!/\text{s}$

4. Matching with options

Thus the angular momentum is $\boxed{6\text{ kg-m}^2/\text{s}}$

So the correct option is A.