1. We are given $f(x)=\frac{5x+3}{6x-\alpha}, \qquad x\neq \frac{\alpha}{6}.$ We need to find $\alpha$ such that $(f\circ f)(x)=x$ for all allowed $x$.

2. Let $y=f(x)=\frac{5x+3}{6x-\alpha}.$ Then $f(f(x))=f(y)=\frac{5y+3}{6y-\alpha}.$ Now substitute $y=\frac{5x+3}{6x-\alpha}$.

3. Compute the numerator: $5y+3=5\left(\frac{5x+3}{6x-\alpha}\right)+3$ $=\frac{25x+15+3(6x-\alpha)}{6x-\alpha}$ $=\frac{25x+15+18x-3\alpha}{6x-\alpha}$ $=\frac{43x+15-3\alpha}{6x-\alpha}.$

4. Compute the denominator: $6y-\alpha=6\left(\frac{5x+3}{6x-\alpha}\right)-\alpha$ $=\frac{30x+18-\alpha(6x-\alpha)}{6x-\alpha}$ $=\frac{30x+18-6\alpha x+\alpha^2}{6x-\alpha}$ $=\frac{(30-6\alpha)x+(18+\alpha^2)}{6x-\alpha}.$

5. Therefore, $f(f(x))=\frac{\frac{43x+15-3\alpha}{6x-\alpha}}{\frac{(30-6\alpha)x+(18+\alpha^2)}{6x-\alpha}}$ $=\frac{43x+15-3\alpha}{(30-6\alpha)x+(18+\alpha^2)}.$ We want this to equal $x$ for all $x$ in the domain: $\frac{43x+15-3\alpha}{(30-6\alpha)x+(18+\alpha^2)}=x.$

6. Cross-multiply: $43x+15-3\alpha=x\big((30-6\alpha)x+(18+\alpha^2)\big).$ So, $43x+15-3\alpha=(30-6\alpha)x^2+(18+\alpha^2)x.$

7. Since this must hold for all $x$, compare coefficients of powers of $x$.

• Coefficient of $x^2$: $30-6\alpha=0 \implies \alpha=5.$

• Coefficient of $x$: $18+\alpha^2=43.$ For $\alpha=5$, $18+25=43,$ which is true.

• Constant term: $15-3\alpha=0.$ For $\alpha=5$, $15-15=0,$ which is also true.

Thus, $\alpha=5$ is the required value.

8. Verify directly for $\alpha=5$: $f(x)=\frac{5x+3}{6x-5}.$ Then $f(f(x))=x,$ so the condition is satisfied.

9. Hence the correct option is: $\boxed{\text{B: }5}.$