First establish that the formula for the capacitor of a parallel plate is:
C = E * (A/d)
where E = permittivity of free space = 8.854*10^-12 C^2/N*m^2
A = area, and d = distance between the plates.
Now also use the formula for capacitance, which is
C = Q/V
Q = charge
V = Voltage or potential difference = 180 V
Rearrange the equation for capacitance so that:
Q = C*V
now plug back in the formulat for parallel plate Capacitor for C, resulting in
Q = E * (A/d) * V
In the question you are given charge density of 29 nC/cm^2, first convert this to SI units.
so = 29 nC/cm^2 = 29*10^-6 C/ (1*10^-2 m)^2
note that charge density is equal to Q/A
so Q/A = 29*10^-6 C/ (1*10^-2 m)^2
So rearrange this equation: Q = E * (A/d) * V, so that Q/A are on the left:
Q/A = (E/d) * V
and so, charge density = (E/d) * V
Now rearrange equation and solve for d.
d = E*V / charge density
Substitute in values and solve for d:
d = (8.854*10^-12 C^2/N*m^2) * (180V) / 29*10^-6 C/ (1*10^-2 m)^2
d = 5.49 * 10 ^-9 m = 5.49 µm