We will start with the understanding that the first derivative of the angular momentum is equivalent to the torque, therefore,
[tex]\frac{d\vec{l}}{dt} = \vec{\tau_{net}}[/tex]
Vectorially the given value can be expressed as,
[tex]\tau_{net} = \vec{\tau_x}+\vec{\tau_y}[/tex]
[tex]\tau_{net} = 2.0\hat{i}+4.0\hat{-j}[/tex]
This will allow us to find the magnitude of the total torque by the trigonometric properties of the vectors, for which:
[tex]|\tau_{net}| = \sqrt{2^2+16^2}[/tex]
[tex]|\tau_{net}| = 5\sqrt{2}N\cdot m[/tex]
Finally the angle will be given by the relation of the tangent,
[tex]tan\theta = \frac{4}{2}[/tex]
[tex]\theta = 63.4\°[/tex]
Direction of 63.4° clockwise from x-direction
