Given ellipse

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

We are told:

• it meets the line $\frac{x}{7}+\frac{y}{2\sqrt6}=1$ on the $x$-axis,
• and the line $\frac{x}{7}-\frac{y}{2\sqrt6}=1$ on the $y$-axis.

Interpret the first condition

The line $\frac{x}{7}+\frac{y}{2\sqrt6}=1$ meets the ellipse at a point on the $x$-axis.

On the $x$-axis, $y=0$. Substituting into the line: $\frac{x}{7}=1 \implies x=7$

So the point $(7,0)$ lies on the ellipse.

Substitute into the ellipse: $\frac{7^2}{a^2}=1 \implies a^2=49$

Interpret the second condition

The line $\frac{x}{7}-\frac{y}{2\sqrt6}=1$ meets the ellipse at a point on the $y$-axis.

On the $y$-axis, $x=0$. Substituting into the line: $-\frac{y}{2\sqrt6}=1 \implies y=-2\sqrt6$

So the point $(0,-2\sqrt6)$ lies on the ellipse.

Substitute into the ellipse: $\frac{(-2\sqrt6)^2}{b^2}=1$ $\frac{24}{b^2}=1 \implies b^2=24$

Find the eccentricity

Since $a^2=49>b^2=24$, the major axis is along the $x$-axis.

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ eccentricity is $e=\sqrt{1-\frac{b^2}{a^2}}$

Therefore, $e=\sqrt{1-\frac{24}{49}}=\sqrt{\frac{25}{49}}=\frac{5}{7}$

Compare with options

$\boxed{\frac{5}{7}}$

So the correct option is A.