1. In a meter bridge at balance point, the ratio of resistances in the two gaps equals the ratio of resistances of the two segments of the bridge wire:

$\frac{R_1}{R_2}=\frac{R_{AC}}{R_{CB}}$

Since the wire is uniform,

$R = \rho \frac{L}{A}$

So for the two segments of the same wire,

$\frac{R_{AC}}{R_{CB}}=\frac{\rho \frac{l}{A}}{\rho \frac{100-l}{A}}=\frac{l}{100-l}$

where $l$ is the balancing length in cm.

2. Initially, the balance point is at

$l=40\text{ cm}$

So,

$\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$

3. Now the radius of the wire $AB$ is doubled.

If radius $r$ is doubled, cross-sectional area becomes

$A' = \pi (2r)^2 = 4\pi r^2 = 4A$

Hence resistance per unit length becomes one-fourth:

$\frac{\rho}{A'}=\frac{1}{4}\frac{\rho}{A}$

But this change happens uniformly throughout the entire bridge wire.

So the resistance of segment $AC$ and segment $CB$ both become one-fourth of their original values:

$R'_{AC}=\frac{1}{4}R_{AC}, \qquad R'_{CB}=\frac{1}{4}R_{CB}$

Therefore their ratio remains unchanged:

$\frac{R'_{AC}}{R'_{CB}}=\frac{R_{AC}}{R_{CB}}$

Thus the balance condition remains at the same length.

4. Therefore, the new balancing length is

$\boxed{40\text{ cm}}$