The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$

Where

I is the impulse

F is the force

$Delta t$ is the time interval during which the force is applied

$\Delta p$ is the change in momentum

In this problem,

$\Delta t = 3.0 s$ is the time interval

$I=6.0 N\cdot s$ (east) is the impulse

Therefore, the magnitude of the force is

$F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N$

And the direction is the same as the impulse (east).

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