A drop of liquid of density ρ is floating half immersed in a liquid...

Balancing the forces on drop

Buoyant force $+$ surface tension $= mg$

$\sigma\dfrac{V}{2}g + 2\piRT = ρVg$

$2\piRT = \dfrac{\left(2ρ - \sigma\right)}{2} \cdot \dfrac{4}{3}\pi{R}^{3}g; \left[\text{As}V = \dfrac{4}{3}\pi{R}^{3}\right]$

$⇒ {R}^{3} = \dfrac{3T}{\left(2ρ - \sigma\right)g} ⇒ R = \sqrt{\dfrac{3 \times 7 . 5 \times {10}^{- 2}N - {m}^{- 1}}{\left(2ρ - \sigma\right) \times 10}}$

$R = \dfrac{3}{20\sqrt{\left(2ρ - \sigma\right)}}m = \dfrac{15}{\sqrt{2ρ - \sigma}}cm$