A wooden block of mass 0.5 kg and density 800 kg m^ -3 is attached...

Volume of the wooden block is $V = \dfrac{m}{\rho_{b}} = \dfrac{0.5}{800} \text{ m}^{3}$

Buoyant force acting on the block in water is $F_{B} = V \rho_{w} g$

Substituting the values, $F_{B} = \left(\dfrac{0.5}{800}\right) \times 1000 \times 10 = 6.25 \text{ N}$

Weight of the block is $W = mg = 0.5 \times 10 = 5 \text{ N}$

Since $F_{B} > W$, the net force on the block is directed upwards. To maintain equilibrium, the spring must exert a downward force, which means it will be elongated.

Let $x$ be the elongation of the spring. At equilibrium, $F_{B} = W + kx$

$6.25 = 5 + 50x$

$50x = 1.25$

$x = \dfrac{1.25}{50} \text{ m} = 0.025 \text{ m}$

$x = 2.5 \text{ cm}$