The software rendering this question really butchered it... I think it's saying the displacement vector for the particle is
[tex]\vec r(t) = \sin(t) \, \vec\imath + 4t^3 \, \vec\jmath + 5t \, \vec k[/tex]
Differentiate twice with respect to time [tex]t[/tex] to get the velocity and acceleration vectors, respectively:
[tex]\dfrac{d\vec r}{dt} = \vec v(t) = \cos(t) \,\vec\imath + 12t^2 \,\vec\jmath + 5 \,\vec k[/tex]
[tex]\dfrac{d^2\vec r}{dt^2} = \dfrac{d\vec v}{dt} = \vec a(t) = -\sin(t) \,\vec\imath + 24t \,\vec\jmath[/tex]
Then at [tex]t=3[/tex], the acceleration of the particle is
[tex]\vec a(3) = -\sin(3)\,\vec\imath + 72\,\vec\jmath[/tex]
which most closely resembles the first choice.
