Answer
$(D)$
Work Step by Step
The major axis is along the $y$-axis, center is at (0,0).
The denominator under $y^{2}$ is going to be larger.
The vertices are $(0,\pm 4),\quad a=4, \quad a^{2}=16$
The minor axis has length $2b=4, $
so $\quad b=2, \quad b^{2}=4$
The equation is$\quad$ $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{16}=1$,
....choice $(D)$