Answer
$F_{max} = 1000~N$
Work Step by Step
The impulse is equal to the area under the force versus time graph. We can find the area under the graph as:
$Area = \frac{1}{2}(F_{max})(2\times 10^{-3}~s)+(F_{max})(4\times 10^{-3}~s)+\frac{1}{2}(F_{max})(2\times 10^{-3}~s)$
$Area = (F_{max})(6\times 10^{-3}~s)$
To find the value of $F_{max}$, we can equate the area under the graph to the impulse.
$(F_{max})(6\times 10^{-3}~s) = 6.0~N~s$
$F_{max}= \frac{6.0~N~s}{(6\times 10^{-3}~s)}$
$F_{max} = 1000~N$