Answer:
4456.33N
Step-by-step explanation:
Let the energy converted by the lifter's body into work done be E.
The work done in lifting the barbell is against gravity and it is given by equation (1);

where m is the mass of the barbell, g is acceleration due to gravity and h is the vertical distance (height) through which it is lifted. Hence by the principle of energy conversion as specified by the problem we could write the following equation;

Also, it should be recalled that the weight, W of an object is given by the equation below;

Substituting (3) into (2) we obtain the following;

Given; h = 1.9m
To obtain the energy E converted by the lifter's body in Joules, we use the specifications given in the problem as follows;
Given; 1cal = 4186J
therefore 113cal = (113 x 4186)J = 473018J. This is the total energy contained in one ounce of cereal consumed by the lifter. However as specified by the problem, only 1.79% of this energy was converted by the lifter's body into weight lifting, therefore;

Putting this value of E into equation (4), we obtain the following;

therefore;

This is the heaviest weight of barbell the lifter could lift with the amount of energy specified.