1. Find the intersection point of the given lines

The lines are $bx+10y-8=0$ and $2x-3y=0.$

From $2x-3y=0$, we get $x=\frac{3y}{2}.$

Substitute into $bx+10y-8=0$: $b\left(\frac{3y}{2}\right)+10y-8=0$ $\left(\frac{3b}{2}+10\right)y=8$ $y=\frac{16}{3b+20}.$

Hence $x=\frac{3}{2}\cdot \frac{16}{3b+20}=\frac{24}{3b+20}.$

So the intersection point is $P\left(\frac{24}{3b+20},\frac{16}{3b+20}\right).$

2. Equation of line through $P$ and $(1,1)$

Since line $L$ passes through $P$ and $(1,1)$, its slope is

Simplify: $m=\frac{\frac{3b+20-16}{3b+20}}{\frac{3b+20-24}{3b+20}}=\frac{3b+4}{3b-4}.$

Thus line $L$ through $(1,1)$ is $y-1=\frac{3b+4}{3b-4}(x-1).$

3. Use tangency condition with the circle

The circle is $17(x^2+y^2)=16 \quad \Rightarrow \quad x^2+y^2=\frac{16}{17}.$

So its center is $(0,0)$ and radius is $r=\frac{4}{\sqrt{17}}.$

Write the line in general form. From $y-1=m(x-1),$ we get $mx-y-m+1=0.$

Distance of the center $(0,0)$ from this line must equal the radius: $\frac{|1-m|}{\sqrt{m^2+1}}=\frac{4}{\sqrt{17}}.$

Now $m=\frac{3b+4}{3b-4}.$

Compute $1-m$: $1-m=1-\frac{3b+4}{3b-4}=\frac{3b-4-(3b+4)}{3b-4}=\frac{-8}{3b-4}.$ So $|1-m|=\frac{8}{|3b-4|}.$

Next, $m^2+1=\frac{(3b+4)^2+(3b-4)^2}{(3b-4)^2}.$

Hence $\sqrt{m^2+1}=\frac{\sqrt{(3b+4)^2+(3b-4)^2}}{|3b-4|}.$

Therefore, $\frac{|1-m|}{\sqrt{m^2+1}}=\frac{8}{\sqrt{(3b+4)^2+(3b-4)^2}}.$

Set this equal to $\frac{4}{\sqrt{17}}$: $\frac{8}{\sqrt{(3b+4)^2+(3b-4)^2}}=\frac{4}{\sqrt{17}}.$

Cross-multiplying, $2\sqrt{17}=\sqrt{(3b+4)^2+(3b-4)^2}.$

Squaring, $68=(3b+4)^2+(3b-4)^2.$

Expand: $(3b+4)^2+(3b-4)^2=(9b^2+24b+16)+(9b^2-24b+16)=18b^2+32.$

So $18b^2+32=68$ $18b^2=36$ $b^2=2.$

4. Find the eccentricity of the ellipse

The ellipse is $\frac{x^2}{5}+\frac{y^2}{b^2}=1.$

Since $b^2=2$, it becomes $\frac{x^2}{5}+\frac{y^2}{2}=1.$

Here, $a^2=5, \quad c^2=a^2-b^2=5-2=3.$

Therefore eccentricity is $e=\frac{c}{a}=\sqrt{\frac{3}{5}}.$

5. Check options

$\sqrt{\frac{3}{5}}$ corresponds to Option B.

Final Answer: $\boxed{\sqrt{\frac{3}{5}}}$