1. Initial geometry

The point of suspension is at height $H=0.9\,\text{m}$ above the floor, while the string length is $L=1.0\,\text{m}$.

Initially, the bob lies on the floor directly below the support. Hence the vertical separation from support to floor is $0.9\,\text{m}$, but the string length is $1.0\,\text{m}$.

So initially the string is slack by $L-H=1.0-0.9=0.1\,\text{m}.$

2. Motion after horizontal impulse

A horizontal impulse $P=0.2\,\text{kg-m/s}$ is given to the bob.

Since the floor is frictionless, while the bob remains on the floor it moves horizontally with constant speed. Its horizontal momentum remains equal to $P$.

Thus the speed of the bob during sliding is $v=\frac{P}{m}=\frac{0.2}{0.1}=2\,\text{m/s}.$

3. Condition when the string becomes taut

Let the support be vertically above the initial point. If the bob has moved a horizontal distance $x$, then the distance from the support to the bob is $\sqrt{H^2+x^2}.$

The string becomes taut when this distance equals $L$: $\sqrt{H^2+x^2}=L.$ So, $x=\sqrt{L^2-H^2}=\sqrt{1^2-0.9^2}=\sqrt{1-0.81}=\sqrt{0.19}.$

4. Angular momentum about the point of suspension just before lift-off

Just before lift-off, the bob is still moving horizontally with momentum magnitude $P=mv=0.2\,\text{kg-m/s}$.

Angular momentum magnitude about the support is $J = r_\perp\, p,$ where $r_\perp$ is the perpendicular distance from the support to the line of motion.

Since the bob is moving horizontally along the floor, the line of motion is a horizontal line at vertical distance $H=0.9\,\text{m}$ below the support. Therefore, $r_\perp = H=0.9\,\text{m}.$

Hence, $J = P\,H = 0.2\times 0.9 = 0.18\,\text{kg-m}^2/\text{s}.$

So, $\boxed{J=0.18}.$

5. Comparison with stored answer

Stored correct answer: $0.18$

Our derived answer matches the stored answer exactly.