Given:
Initial radius of Earth, \( r_1 \)
Final radius of Earth, \( r_2 = \frac{1}{2} \times r_1 \) (since the diameter becomes half)
Mass of Earth, \( M \) (constant)

The acceleration due to gravity (\( g \)) at the surface of Earth is given by:

\[ g = \frac{G \cdot M}{r^2} \]

where:
\( G \) is the gravitational constant,
\( M \) is the mass of Earth, and
\( r \) is the radius of Earth.

Let’s denote the initial acceleration due to gravity as \( g_1 \) and the final acceleration due to gravity as \( g_2 \).

So, initially, \( g_1 = \frac{G \cdot M}{r_1^2} \).

When the diameter of Earth becomes half, the new radius becomes \( r_2 = \frac{1}{2} \times r_1 \).

Therefore, the final acceleration due to gravity becomes:

\[ g_2 = \frac{G \cdot M}{(r_1/2)^2} = \frac{G \cdot M}{\frac{1}{4} \cdot r_1^2} = \frac{4 \cdot G \cdot M}{r_1^2} = 4 \cdot g_1 \]

So, the acceleration due to gravity at the surface of Earth becomes four times its initial value.

Therefore, the correct option is (b) Four times.