1. Write the standard equation of the ellipse

Since the ellipse has centre at the origin and axes along the coordinate axes, its equation is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

where $a,b>0$.

2. Use the given points

The ellipse passes through $(4,-1)$ and $(-2,2)$.

So these satisfy the equation:

For $(4,-1)$:

$\frac{16}{a^2}+\frac{1}{b^2}=1 \quad ...(1)$

For $(-2,2)$:

$\frac{4}{a^2}+\frac{4}{b^2}=1 \quad ...(2)$

3. Substitute

Let

$u=\frac{1}{a^2}, \qquad v=\frac{1}{b^2}$

Then equations (1) and (2) become:

$16u+v=1 \quad ...(3)$ $4u+4v=1 \quad ...(4)$

From (4):

$u+v=\frac14 \quad ...(5)$

Now subtract (5) from (3):

$15u=\frac34$

$u=\frac{1}{20}$

So,

$a^2=20$

From (5):

$v=\frac14-\frac{1}{20}=\frac{5-1}{20}=\frac{4}{20}=\frac15$

Hence,

$b^2=5$

Thus the ellipse is

$\frac{x^2}{20}+\frac{y^2}{5}=1$

4. Find the eccentricity

Since $a^2=20>b^2=5$, the eccentricity is

$e=\sqrt{1-\frac{b^2}{a^2}}$

$e=\sqrt{1-\frac{5}{20}}=\sqrt{1-\frac14}=\sqrt{\frac34}=\frac{\sqrt3}{2}$

5. Match with the options

$\frac{\sqrt3}{2}$ corresponds to Option C.

Therefore, the eccentricity is

$\boxed{\frac{\sqrt3}{2}}$