Given,
The range of the flight of the arrow, R=30.0 m
The initial launch velocity of the arrow, u=110.0 m
It is given that the arrow is launched horizontally, thus the vertical component of the launch velocity of the arrow is 0 m/s.
The range of the flight of the arrow is given by,

Where t is the time of flight of the arrow.
On substituting the known values,

From the equation of motion, the drop in the height of the arrow during its entire is given by,

Where g is the acceleration due to gravity and u_y is the vertical component of the initial velocity of the arrow.
On substituting the known values,

Thus the arrow will hit 66 m below the bullseye.