A small block B is positioned on another block A of mass 5 kg and...

Since all surfaces are frictionless, there is no horizontal force acting on block $B$. Therefore, the acceleration of block $B$ is zero ($a_B = 0$).

The horizontal force $F = 10\text{ N}$ acts only on block $A$. The acceleration of block $A$ is given by:
$a_A = \dfrac{F}{m_A} = \dfrac{10}{5} = 2\text{ m/s}^2$

The relative acceleration of block $B$ with respect to block $A$ is:
$a_{rel} = |a_B - a_A| = |0 - 2| = 2\text{ m/s}^2$

Block $B$ is initially at the right end of block $A$. To separate from block $A$, it must cover a relative distance equal to the length of block $A$:
$s_{rel} = 20\text{ cm} = 0.2\text{ m}$

Using the second equation of motion for relative motion ($u_{rel} = 0$):
$s_{rel} = \dfrac{1}{2} a_{rel} t^2$

Substituting the values:
$0.2 = \dfrac{1}{2} \times 2 \times t^2$

$t^2 = 0.2$

$t = \sqrt{0.2} \approx 0.447\text{ s}$

Rounding to two decimal places, the time elapsed is $0.45\text{ s}$.