Use Hooke's law... (just kidding)
Break down each force vector into horizontal and vertical components.
[tex]\vec F_1=(1000\,\mathrm N)(\cos30^\circ\,\vec x+\sin30^\circ\,\vec y)\approx(866.025\,\mathrm N)\,\vec x+(500\,\mathrm N)\,\vec y[/tex]
[tex]\vec F_2=(1500\,\mathrm N)(\cos160^\circ\,\vec x+\sin160^\circ\,\vec y)\approx(-1409.54\,\mathrm N)\,\vec x+(513.03\,\mathrm N)\,\vec y[/tex]
[tex]\vec F_3=(750\,\mathrm N)(\cos195^\circ\,\vec x+\sin195^\circ\,\vec y)\approx(-724.444\,\mathrm N)\,\vec x+(-194.114\,\mathrm N)\,\vec y[/tex]
The resultant force is the sum of these vectors,
[tex]\vec F=\displaystyle\sum_{i=1}^3\vec F_i\approx(-1267.96\,\mathrm N)\,\vec x+(818.916\,\mathrm N)\,\vec y[/tex]
and has magnitude
[tex]|\vec F|\approx\sqrt{(-1267.96\,\mathrm N)^2+(818.916\,\mathrm N)^2}\approx1509.42\,\mathrm N[/tex]
The closest answer is D.
