1. Let $|\vec a|=A,\quad |\vec b|=B$ and the angle between $\vec a$ and $\vec b$ be $60^\circ$.

2. Given that $\dfrac{1}{8}\vec a$ is a unit vector, so $\left|\frac{1}{8}\vec a\right|=1 \implies \frac{1}{8}|\vec a|=1 \implies A=8.$

3. Use the condition $|2\vec a+3\vec b|=|3\vec a+\vec b|.$ Squaring both sides, $|2\vec a+3\vec b|^2=|3\vec a+\vec b|^2.$

4. Expand both sides using $|\vec x+\vec y|^2=|\vec x|^2+|\vec y|^2+2\vec x\cdot\vec y.$

   Left side: $|2\vec a+3\vec b|^2=4A^2+9B^2+12\vec a\cdot\vec b.$

   Right side: $|3\vec a+\vec b|^2=9A^2+B^2+6\vec a\cdot\vec b.$

5. Since angle between $\vec a$ and $\vec b$ is $60^\circ$, $\vec a\cdot\vec b=AB\cos 60^\circ=\frac{AB}{2}.$

   Substitute: $4A^2+9B^2+12\cdot \frac{AB}{2}=9A^2+B^2+6\cdot \frac{AB}{2}.$

   So, $4A^2+9B^2+6AB=9A^2+B^2+3AB.$

6. Rearranging, $8B^2+3AB-5A^2=0.$

7. Now put $A=8$: $8B^2+3(8)B-5(64)=0$ $8B^2+24B-320=0$ Divide by $8$: $B^2+3B-40=0$ $(B+8)(B-5)=0.$

8. Since magnitude is positive, $B=5.$

Therefore, $|\vec b|=5.$

So the correct option is C.