A thin disc of radius b=2a has a concentric hole of radius 'a' in it. It carries uniform surface charge 'σ' on it. If the electric field on its axis at height 'h' (h<<a) from its centre is given as 'Ch', the value of 'C' is.

The electric field due to a uniform charge disk with surface charge density σ is given as
E=σ2ε₀[1−h√h²+R²]
where h is the distance along the axis of the disk from the center of the disk.

The electric field at distance h from the center for the given distribution can be found by subtracting the electric field due to a disk of radius a from the disk of radius 2a:
Enet=Edisk radius 2a − Edisk radius a
E=σ2ε₀(1−h√h²+(2a)² )−σ2ε₀(1−h√h²+a²). (i)

as h<<a, √h²+(2a)² ≈2a and √h²+(a)² ≈a
Substituting in equation (i):
E=σ2ε₀(1−h2a)−σ2ε₀(1−ha)
E=σ2ε₀h2a
E=σh4ε₀a
E=σ4ε₀ah
The value of C is
C=σ4ε₀a
Hence option C is correct