$I_{\text {support }}=m R^{2}+m R^{2}=2 m R^{2}$

$T=2 \pi \sqrt{\frac{\left(2 m R^{2}\right)}{m g R}} =2 \pi \sqrt{\frac{2 R}{g}}$

$\Rightarrow R=\frac{T^{2} g}{8 \pi^{2}}$

A seconds pendulum is a pendulum whose period is precisely $two\, seconds; one \,second$ for a swing in one direction and $one\, second$ for the return swing. Ie. $T=2 \varepsilon,$ so we have

$R=\frac{2^{2} \times \pi^{2}}{8 \pi^{2}}=0.5 \mathrm{m}$