Answer:
The electric field strength inside the capacitor is 49880.77 N/C.
Explanation:
Given:
Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m
Separation between the plates (d) = 0.407 mm = 
Energy stored in the capacitor (U) = 
Assuming the medium to be air.
So, permittivity of space (ε) = 
Area of the square plates is given as:

Capacitance of the capacitor is given as:

Now, we know that, the energy stored in a parallel plate capacitor is given as:

Rewriting in terms of 'E', we get:

Now, plug in the given values and solve for 'E'. This gives,

Therefore, the electric field strength inside the capacitor is 49880.77 N/C