= Solution
Measure $x$ from the clamped end and let downward deflection $y$ be positive. In the small-slope <elastic beam> model, the <bending moment> has magnitude $M(x)=F(l-x)$ and $EIy''=M(x)$. Integrating with $y(0)=y'(0)=0$ gives
$$
y'(x)=\frac F{EI}\left(lx-\frac{x^2}{2}\right),\qquad y(x)=\frac F{EI}\left(\frac{lx^2}{2}-\frac{x^3}{6}\right).
$$
For the rectangular cross-section, $I=wh^3/12=6.75\times10^{-7}\,\mathrm{m^4}$, and therefore
$$
\boxed{\delta=y(l)=\frac{Fl^3}{3EI}=\frac{4mgl^3}{Ewh^3}.}
$$
Requiring $\delta\le0.2\,\mathrm m$ with $F=1079.1\,\mathrm N$ yields
$$
E\ge\frac{4(110)(9.81)(3)^3}{(0.3)(0.03)^3(0.2)}=\boxed{71.94\,\mathrm{GPa}.}
$$
This calculation neglects beam self-weight as stipulated, and also transverse shear deformation and geometric nonlinearities. The end slope at the limit is $3\delta/(2l)=0.10$, so the small-slope approximation is reasonable for the intended estimate. Meeting a stiffness specification alone does not establish safe strength or fatigue life.
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