Given, density of bob, $\rho =\frac{4}{3}\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$
Density of water, $\sigma ={10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$

If $g^{\text{'}}$ be gravitational acceleration in water then,

$g^{\text{'}}=g \left(1-\frac{\sigma }{\rho }\right)=g \left(1-\frac{{10}^3}{\frac{4}{3}\times {10}^3}\right)=\frac{g}{4}$

As, $T_{\mathrm{air} }=2\pi \sqrt{\frac{I}{g}}$,

Similarly, $T_{\mathrm{water} }=2\pi \sqrt{\frac{I}{g^{\text{'}}}} \left[T_{\mathrm{water} }=2 \mathrm{s}\right] \left(g^{\text{'}}=\frac{g}{4}\right)$

$2=2\pi \sqrt{\frac{I}{g/4}}=2\pi \sqrt{\frac{I}{g}} ·2$

$2=2.T_{\mathrm{air} }$

$\Rightarrow T_{\mathrm{air} }=1 \mathrm{s}$