1. Given data

• Small block: $m = 0.5\,\text{kg}$
• Large block: $M = 4.5\,\text{kg}$
• Coefficient of static friction between blocks: $\mu_s = \dfrac{3}{7}$
• Table is frictionless
• $g = 9.8\,\text{m s}^{-2}$

We need the maximum horizontal force on the larger block such that both blocks move together.

2. Condition for moving together

Since the table is frictionless, the only external horizontal force on the two-block system is $F$.

So if the blocks move together, their common acceleration is

$a = \frac{F}{M+m}$

For the top/smaller block to move with the larger block without slipping, static friction must be sufficient to accelerate it.

Required friction on the small block:

$f = ma$

Maximum available static friction:

$f_{\max} = \mu_s N = \mu_s mg$

For no slipping,

$ma \le \mu_s mg$

Cancelling $m$,

$a \le \mu_s g$

Thus the maximum allowed common acceleration is

$a_{\max} = \mu_s g = \frac{3}{7}\times 9.8 = 4.2\,\text{m s}^{-2}$

3. Find maximum force

Now,

$F_{\max} = (M+m)a_{\max}$

$F_{\max} = (4.5+0.5)\times 4.2 = 5\times 4.2 = 21\,\text{N}$

4. Final answer

$\boxed{21\,\text{N}}$

Rounded to the nearest integer: $\boxed{21}$.

5. Comparison with stored answer

Stored correct answer = $21$

Our derived answer also = $21$, so they agree.