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Take $h$ positive downwards and write $B$ for the filament bending modulus. Under the material-modulus interpretation of the printed $A$ as Young's modulus, the circular cross-section gives the second moment of area and bending modulusIndeed a small curvature $\kappa$ gives axial strain $-y\kappa$, so integrating $A{y}^2{\kappa }^2/2$ over the section gives bending energy $B{\kappa }^2/2$ per unit length. Keeping this distinction is essential to the powers of radius in the answer.

After allowing for buoyancy, the effective downward force per unit length isThe filament loses potential energy as it moves down; the displaced fluid's hydrostatic contribution is included through the density difference. For an inextensible filament with arc-length coordinate $s$, the geometric bending and gravitational energy is ${\int }_0^L[B\kappa (s{)}^2/2-wh(s)]ds$, up to a constant. For the small-slope Monge representation this reduces toThe upward-positive convention reverses the sign of $h$ and the gravitational term together. There is no stretching contribution at this order when bending a slender filament without imposed axial tension. The wording does not explicitly distinguish a material elastic modulus from a flexural modulus: if $A$ instead denotes the filament bending modulus in the intended convention, set $B=A$ throughout; then the $\pi a^4/4$ factor is already included in $A$.