If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original keeping amplitude same then increase in its total energy at a place will be

The total energy of an oscillating simple pendulum is $E = \frac{1}{2} m \omega^2 A^2$ where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.

For a simple pendulum, $\omega = \sqrt{\frac{g}{L}}$ where $g$ is gravitational acceleration and $L$ is length.

Substituting, $E = \frac{1}{2} m \left(\sqrt{\frac{g}{L}}\right)^2 A^2 = \frac{1}{2} \frac{m g A^2}{L}$.

With original length $L_1$, the energy is $E_1 = \frac{1}{2} \frac{m g A^2}{L_1}$.

When the length is reduced to $L_2 = \frac{L_1}{3}$ with amplitude unchanged, the new energy becomes $E_2 = \frac{1}{2} \frac{m g A^2}{L_2} = \frac{1}{2} \frac{m g A^2}{L_1/3} = \frac{3}{2} \frac{m g A^2}{L_1} = 3E_1$.

The increase in total energy is $E_2 - E_1 = 3E_1 - E_1 = 2E_1$.

Final answer: $\boxed{\text{B}}$