To determine the change in magnetic field intensity at the center of a circular loop when its radius is reduced to half and the current is doubled, we need to use the formula for the magnetic field intensity at the center of a current-carrying loop.

The magnetic field B at the center of a circular loop carrying a current I with radius r is given by:

B = \frac{\mu_0 I}{2r}

Here, \mu_0 is the permeability of free space (a constant).

Initially, the magnetic field intensity B_1 is:

B_1 = \frac{\mu_0 I}{2r}

When the radius of the loop is reduced to half and the current is doubled, the new radius becomes \frac{r}{2} and the new current becomes 2I.

The new magnetic field intensity B_2 can be calculated as:

B_2 = \frac{\mu_0 (2I)}{2(\frac{r}{2})}

Simplifying this expression, we get:

B_2 = \frac{\mu_0 \cdot 2I}{r}

Further simplification gives:

B_2 = 4 \times \frac{\mu_0 I}{2r}

Thus:

B_2 = 4 \cdot B_1

Hence, the magnetic field intensity at the center becomes four times the original value when the radius is halved and the current is doubled.

Therefore, the correct answer is: Four times the original value.