#### Answer

The surface charge density will be $2~nC/cm^2$

#### Work Step by Step

We can write an expression for the initial surface charge density:
$\sigma_1 = \frac{q}{A_1}$
If the radius of the disk is doubled, then the area of the disk increases by a factor of 4. That is: $A_2 = 4~A_1$
We can find the final surface charge density:
$\sigma_2 = \frac{q}{A_2}$
$\sigma_2 = \frac{q}{4~A_1}$
$\sigma_2 = \frac{\sigma_1}{4}$
$\sigma_2 = \frac{8~nC/cm^2}{4}$
$\sigma_2 = 2~nC/cm^2$
The surface charge density will be $2~nC/cm^2$.