When bob is immersed in liquid

${mg}_{{eff}}={mg}-$ Buoyant force

${mg}_{{eff}} ={mg}-{v} \sigma {g} \quad(\sigma=\text { density of liquid })$

$={mg}-{v} \frac{\rho}{4} {g}$

$={mg}-\frac{{mg}}{4}=\frac{3 {mg}}{4}$

$\therefore {g}_{{eff}} =\frac{3 {g}}{4}$

${T}_{1}= 2 \pi \sqrt{\frac{\ell_{1}}{{g}_{{eff}}}} \quad \ell_{1}=\ell+\frac{\ell}{3}=\frac{4 \ell}{3}, \quad \ell_{{eff}}=\frac{3 {g}}{4}$

By solving

${T}_{1}=\frac{4}{3} 2 \pi \sqrt{\ell / {g}}$

${T}_{1}=\frac{4 {T}}{3}$