The time period of the disc is $2 \pi \sqrt{(3 r / 2 g )}$

We know that the time period of an object,

$T =2 \pi \sqrt{(I / mgL )}$

where,

$I=$ moment of inertia from the suspended point

$L =$ distance of its centre from suspended point $= r$

we know that, the moment of inertia of disc about its centre $= mr ^2 / 2$

using parallel axis theoram the moment of inertia from a point in its periphery,

$I = mr ^2+ mr ^2 / 2=3 mr ^2 / 2$

putting the values in the above equation we get,

$2 \pi \sqrt{\left(3 mr ^2 / 2 mgr \right)}$

$=2 \pi \sqrt{(3 r / 2 g )}$

therefore, the time period of the disc is $2 \pi \sqrt{(3 r / 2 g )}$