1. We need to find $\cot\left(\frac{\pi}{24}\right)=\cot 7.5^\circ.$

2. Use the half-angle identity: $\cot\frac{\theta}{2}=\csc\theta+\cot\theta.$ Take $\theta=\frac{\pi}{12}=15^\circ$. Then $\cot\left(\frac{\pi}{24}\right)=\csc\left(\frac{\pi}{12}\right)+\cot\left(\frac{\pi}{12}\right).$

So we need $\sin 15^\circ$ and $\cos 15^\circ$.

3. Compute $\sin 15^\circ$ and $\cos 15^\circ$ using angle subtraction: $\sin 15^\circ=\sin(45^\circ-30^\circ)$ $=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ$ $=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\cdot\frac12$ $=\frac{\sqrt3-1}{2\sqrt2}.$

Similarly, $\cos 15^\circ=\cos(45^\circ-30^\circ)$ $=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ$ $=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12$ $=\frac{\sqrt3+1}{2\sqrt2}.$

4. Now compute $\csc 15^\circ$: $\csc 15^\circ=\frac{1}{\sin15^\circ}=\frac{2\sqrt2}{\sqrt3-1}.$ Rationalizing, $\csc 15^\circ=\frac{2\sqrt2(\sqrt3+1)}{3-1}=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2.$

5. Compute $\cot 15^\circ$: $\cot15^\circ=\frac{\cos15^\circ}{\sin15^\circ}=\frac{\sqrt3+1}{\sqrt3-1}.$ Rationalizing, $\cot15^\circ=\frac{(\sqrt3+1)^2}{3-1}$ $=\frac{3+1+2\sqrt3}{2}=2+\sqrt3.$

6. Therefore, $\cot\left(\frac{\pi}{24}\right)=\csc15^\circ+\cot15^\circ$ $=(\sqrt6+\sqrt2)+(2+\sqrt3)$ $=\sqrt2+\sqrt3+2+\sqrt6.$

7. Compare with options:

• A: $\sqrt 2 + \sqrt 3 + 2 - \sqrt 6$
• B: $\sqrt 2 + \sqrt 3 + 2 + \sqrt 6$
• C: $\sqrt 2 - \sqrt 3 - 2 + \sqrt 6$
• D: $3\sqrt 2 - \sqrt 3 - \sqrt 6$

Hence the correct option is $\boxed{\text{B}}.$