1. Given circle

The circle is $x^2+y^2-3x+10y-15=0.$

Write it in the form $x^2+y^2+2gx+2fy+c=0,$ so $2g=-3\Rightarrow g=-\frac32,\qquad 2f=10\Rightarrow f=5,\qquad c=-15.$

Hence its centre is $O=(-g,-f)=\left(\frac32,-5\right).$

2. Equation of tangent to the circle

For the circle $x^2+y^2+2gx+2fy+c=0,$ the tangent at point $(x_1,y_1)$ on the circle is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0.$

3. Tangent at $A(4,-11)$

Using $g=-\frac32, f=5, c=-15$: $4x-11y-\frac32(x+4)+5(y-11)-15=0.$ Simplify: $4x-11y-\frac32x-6+5y-55-15=0$ $\frac52x-6y-76=0.$ Multiplying by $2$: 5x-12y-152=0.\tag{1}

4. Tangent at $B(8,-5)$

Similarly, $8x-5y-\frac32(x+8)+5(y-5)-15=0.$ Simplify: $8x-5y-\frac32x-12+5y-25-15=0$ $\frac{13}{2}x-52=0.$ So, 13x-104=0\Rightarrow x=8.\tag{2}

5. Point of intersection of tangents: $C$

From (2), $x=8.$ Substitute into (1): $5(8)-12y-152=0$ $40-12y-152=0$ $-12y=112$ $y=-\frac{28}{3}.$

Thus, $C=\left(8,-\frac{28}{3}\right).$

6. Required radius

We need the radius of the circle centered at $C$ and tangent to line $AB$. So the radius equals the perpendicular distance from $C$ to the line through $A$ and $B$.

First find equation of line $AB$.

Slope of $AB$: $m=\frac{-5-(-11)}{8-4}=\frac{6}{4}=\frac32.$

Equation through $A(4,-11)$: $y+11=\frac32(x-4).$

So, $2y+22=3x-12$ 3x-2y-34=0.\tag{3}

Distance from $C\left(8,-\frac{28}{3}\right)$ to line (3): $r=\frac{|3(8)-2\left(-\frac{28}{3}\right)-34|}{\sqrt{3^2+(-2)^2}}.$

Compute numerator: $24+\frac{56}{3}-34=\frac{72+56-102}{3}=\frac{26}{3}.$

Denominator: $\sqrt{9+4}=\sqrt{13}.$

Hence, $r=\frac{\frac{26}{3}}{\sqrt{13}}=\frac{26}{3\sqrt{13}}=\frac{2\sqrt{13}}{3}.$

7. Option check

Thus the required radius is $\boxed{\frac{2\sqrt{13}}{3}}.$

So the correct option is A.