The correct answer is $\boxed{\frac{{100}}{{{\text{E}}{I_{\text{c}}}}}}$.
The horizontal deflection of a parabolic curved beam of span 10 m and rise 3 m when loaded with a uniformly distributed load $l$ t per horizontal length is given by the following formula:
$$\delta = \frac{{50}}{{{\text{E}}{I_{\text{c}}}}}$$
where $E$ is the Young’s modulus of the beam material and $I_c$ is the moment of inertia of the beam cross-section at the crown.
The moment of inertia of a parabolic beam cross-section at the crown is given by the following formula:
$$I_c = \frac{{1}{4}}{{b^3}}$$
where $b$ is the width of the beam.
The Young’s modulus of steel is approximately $200 \times {10^9}$ Pa.
The width of a typical steel beam is approximately 0.3 m.
Therefore, the horizontal deflection of a parabolic curved beam of span 10 m and rise 3 m when loaded with a uniformly distributed load $l$ t per horizontal length is approximately:
$$\delta = \frac{{50}}{{{\text{E}}{I_{\text{c}}}}} = \frac{{50}}{{200 \times {10^9} \times \frac{{1}{4}}{{0.3^3}}}} = \frac{{100}}{{{\text{E}}{I_{\text{c}}}}}$$