First find the time it takes for the ball to reach the ground using the vertical component of its position vector:
[tex]y=y_0+v_{0y}t+\dfrac12a_yt^2[/tex]
[tex]\implies0=100\,\mathrm m+\dfrac12\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2[/tex]
[tex]\implies t=4.52\,\mathrm s[/tex]
Meanwhile, the horizontal component of the ball's position vector is
[tex]x=x_0+v_{0x}t+\dfrac12a_xt^2[/tex]
[tex]\implies x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)t[/tex]
After about 4.52 s, the ball has traveled a horizontal distance of
[tex]x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)(4.52\,\mathrm s)=180.8\,\mathrm m[/tex]
which you would round to 200 m, so the answer is B.