#### Answer

$F_{max} = 1500~N$

#### 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:
$Area = \frac{1}{2}(F_{max})(8\times 10^{-3}~s)$
$Area = (F_{max})(4\times 10^{-3}~s)$
To find the value of $F_{max}$, we can equate the area under the graph to the impulse.
$(F_{max})(4\times 10^{-3}~s) = 6.0~N~s$
$F_{max}= \frac{6.0~N~s}{(4\times 10^{-3}~s)}$
$F_{max} = 1500~N$