A

Given:
Incident angle, \( i = 60^\circ \)
Refractive index, \( \mu = 1.5 \)
Thickness of the glass slab, \( d = 10 \text{ cm} = 0.1 \text{ m} \)

Step 1: Calculate the refracted angle using Snell's Law:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
Where \( \mu_1 = 1 \) (air) and \( \mu_2 = 1.5 \).
Thus,
\[ 1 \cdot \sin(60^\circ) = 1.5 \cdot \sin(r) \]
\[ \sin(r) = \frac{\sin(60^\circ)}{1.5} = \frac{\frac{\sqrt{3}}{2}}{1.5} = \frac{\sqrt{3}}{3} \]
This gives us \( r \approx 35.0^\circ \).

Step 2: Calculate the lateral displacement (x) using the formula:
\[ x = d \cdot \sin(i - r) \]
Substituting the values,
\[ x = 0.1 \cdot \sin(60^\circ - 35^\circ) \]
\[ x = 0.1 \cdot \sin(25^\circ) \approx 0.1 \cdot 0.4226 \approx 0.04226 \text{ m} = 4.226 ext{ cm} \]

Therefore, the lateral displacement is approximately \( 4.23 ext{ cm} \), and the correct option is A.