1. Use conservation of mechanical energy
   Since the path is frictionless and the body starts from rest, loss in potential energy equals gain in kinetic energy.

$m g \Delta h = \frac12 m v_B^2$

So,

$v_B = \sqrt{2 g \Delta h}$

2. Find the vertical drop from $A$ to $B$
   From the figure description, the body moves along a quarter of a vertical circle from the top-side point $A$ to the bottom point $B$.
   Hence the vertical height difference is

$\Delta h = R + R = 2R?$

But for the standard geometry of such a question, point $A$ is the side point of the circle and $B$ is the lowest point, so the drop is actually

$\Delta h = R$

That would give

$v_B = \sqrt{2gR} = \sqrt{2 \times 10 \times 14} = \sqrt{280}$

Using $\sqrt{2}=1.4$,

$\sqrt{280} = \sqrt{2 \times 140} = 1.4\sqrt{140}$

This is about $16.7\,\text{m/s}$, which matches option C.

However, the stored answer is D, so let us check the other likely geometry.

3. If $A$ is the top point and $B$ is the bottom point of the vertical circle
   Then the total vertical drop is the diameter:

$\Delta h = 2R = 28\,\text{m}$

Now,

$m g (2R) = \frac12 m v_B^2$

$v_B = \sqrt{4gR} = 2\sqrt{gR}$

Substitute $g=10$ and $R=14$:

$v_B = \sqrt{4\times 10 \times 14} = \sqrt{560}$

$\sqrt{560} = \sqrt{56\times 10} = \sqrt{4\times 14\times 10}$

Numerically,

$\sqrt{560} \approx 23.7\,\text{m/s}$

This does not match any option.

4. Check option D numerically
   Option D is $21.9\,\text{m/s}$. Its square is approximately

$21.9^2 \approx 479.6$

So the corresponding height drop would be

$\Delta h = \frac{v^2}{2g} \approx \frac{479.6}{20} \approx 24\,\text{m}$

That is neither $R=14$ nor $2R=28$.

5. Conclusion
   For the usual interpretation of the given circular path figure in such problems, the body drops through height $R$, giving

$v_B = \sqrt{2gR} = \sqrt{280} \approx 16.7\,\text{m/s}$

So the correct option should be:

$\boxed{\text{C: }16.7\,\text{m/s}}$