1. Given hyperbola

   $\frac{x^2}{9}-\frac{y^2}{b^2}=1$

   Comparing with the standard form

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

   we get

   $a^2=9 \implies a=3.$

2. Endpoints of a latus rectum

   For the hyperbola

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

   we have

   $c^2=a^2+b^2.$

   The latus rectum through the focus $(c,0)$ is the line $x=c$.

   Substituting $x=c$ into the hyperbola:

   $\frac{c^2}{a^2}-\frac{y^2}{b^2}=1.$

   Since $c^2=a^2+b^2$,

   $\frac{a^2+b^2}{a^2}-\frac{y^2}{b^2}=1$ $1+\frac{b^2}{a^2}-\frac{y^2}{b^2}=1$ $\frac{y^2}{b^2}=\frac{b^2}{a^2}$ $y=\pm \frac{b^2}{a}.$

   So the endpoints of the latus rectum are

   $(c,\, b^2/a), \quad (c,\, -b^2/a).$

3. Angle subtended at the centre

   The centre is the origin $(0,0)$. Let the endpoints be

   $P\left(c,\frac{b^2}{a}\right), \qquad Q\left(c,-\frac{b^2}{a}\right).$

   The angle $\angle POQ=\frac{\pi}{3}$.

   Using the dot product:

   $\cos \angle POQ = \frac{\vec{OP}\cdot \vec{OQ}}{|OP|\,|OQ|}.$

   Now,

   $\vec{OP}=\left(c,\frac{b^2}{a}\right), \qquad \vec{OQ}=\left(c,-\frac{b^2}{a}\right).$

   Therefore,

   $\vec{OP}\cdot \vec{OQ}=c^2-\frac{b^4}{a^2}.$

   Also,

   $|OP|=|OQ|=\sqrt{c^2+\frac{b^4}{a^2}}.$

   Hence,

   $\cos\frac{\pi}{3}=\frac{c^2-\frac{b^4}{a^2}}{c^2+\frac{b^4}{a^2}}.$

   Since $\cos\frac{\pi}{3}=\frac12$,

   $\frac{c^2-\frac{b^4}{a^2}}{c^2+\frac{b^4}{a^2}}=\frac12.$

4. Solve for $b^2$

   Let

   $t=\frac{b^4}{a^2}.$

   Then

   $\frac{c^2-t}{c^2+t}=\frac12$ $2(c^2-t)=c^2+t$ $2c^2-2t=c^2+t$ $c^2=3t.$

   So,

   $a^2+b^2 = 3\cdot \frac{b^4}{a^2}.$

   Since $a^2=9$,

   $9+b^2=3\cdot \frac{b^4}{9}=\frac{b^4}{3}.$

   Multiply by $3$:

   $27+3b^2=b^4.$

   Let $u=b^2$. Then

   $u^2-3u-27=0.$

   Solving,

   $u=\frac{3\pm\sqrt{9+108}}{2}=\frac{3\pm\sqrt{117}}{2}=\frac{3\pm 3\sqrt{13}}{2}.$

   Since $b^2>0$,

   $b^2=\frac{3}{2}(1+\sqrt{13}).$

5. Identify $l,m,n$

   Comparing with

   $b^2=\frac{l}{m}(1+\sqrt{n}),$

   we get

   $l=3,\quad m=2,\quad n=13.$

   Therefore,

   $l^2+m^2+n^2=3^2+2^2+13^2=9+4+169=182.$

6. Final answer

   $\boxed{182}$

The derived answer matches the stored correct answer.