The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :

$$ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} $$

After the collision, the bullet and the bob move together with a common velocity. Let's denote this common velocity as $ v' $. The final momentum $ p_{\text{final}} $ is the combined mass of the bullet and bob times the common velocity :

$$ p_{\text{final}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $$

According to the principle of conservation of linear momentum,

$$ p_{\text{initial}} = p_{\text{final}} $$

$$ m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{bob}}) \times v' $$

Plugging in the values :

$$ (10^{-2} \text{ kg}) \times (2 \times 10^2 \text{ m/s}) = (10^{-2} \text{ kg} + 1 \text{ kg}) \times v' $$

Solving for $ v' $ :

$$ v' = \frac{10^{-2} \times 2 \times 10^2}{10^{-2} + 1} = \frac{2}{1.01} \approx 1.98 \text{ m/s} $$

After the collision, the system has some kinetic energy which will be completely converted to potential energy at the maximum height $ h $ that the bob reaches. Using the principle of conservation of energy :

$$ \text{Kinetic Energy (KE)}_{\text{initial}} = \text{Potential Energy (PE)}_{\text{final}} $$

$$ \frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2 = (m_{\text{bullet}} + m_{\text{bob}})gh $$

Isolating $ h $, we get :

$$ h = \frac{\frac{1}{2}(m_{\text{bullet}} + m_{\text{bob}})v'^2}{(m_{\text{bullet}} + m_{\text{bob}})g} = \frac{v'^2}{2g} $$

Substituting the values for $ v' $ and $ g $ :

$$ h = \frac{(1.98)^2}{2 \times 10} = \frac{3.9204}{20} = 0.19602 \text{ m} $$

Looking at the given options, the result most closely matches Option A :

$$ \boxed{0.20 \text{ m}} $$

Therefore, the height to which the bob rises before swinging back is approximately $ 0.20 \text{ m} $.