Attempt:

$y_{COM} = \dfrac{R}{3} = \dfrac {25}3$

$I = \dfrac 12 (2)R^2 + 1R^2 = 2\times 25 = 50 $

Now, T is given by: $T =2\pi \sqrt{\dfrac{I}{mgl}}$ where l is the distance of the centre of mass from the centre of rotation

So $T =2\pi \sqrt{\dfrac{50}{3 \times 10 \times \dfrac{25}{3}}} = 2\pi\sqrt{\dfrac{1}{5}}$

No option :(