We have an expression:
\begin{equation} B = \frac{FL^{3}}{3EI} = \frac{N m^{3}}{3 p m^{4}} = \frac{Nm^{3}}{\frac{N}{m^{2}}m^{4}} = m \end{equation}
With constants: B: m, deflection at the point of force application F: N, force applied L: m, distance between fixed point and point of force application E: p=\frac{N}{m^{2}}, elastic modulus I: m^{4}, second moment of area As per measured: B: 9.15 \cdot 10^{-4} m F: 20N L: 9.373 \cdot 10^{-2} m I: 1.37 \cdot 10^{-10} m^{4} = \frac{WH^{3}}{12} = \frac{(6.25 \cdot 10^{-3})(6.4 \cdot 10^{-3})^{3}}{12} Theoretical: E: 7 \cdot 10^{10} P As calculated: B: 5.74 \cdot 10^{-4} m