In a solvent S , a compound B is partially dissociated into C and D...

The elevation in boiling point is given by $\Delta T_b = i K_b m$.

The ebullioscopic constant $K_b$ is calculated as:
$K_b = \dfrac{R (T_b^\circ)^2 M_S}{1000 \Delta H_{vap}}$

Substituting the given values ($T_b^\circ = 400$ K, $\Delta H_{vap} = 10R$ J mol$^{-1}$):
$K_b = \dfrac{R (400)^2 M_S}{1000 (10R)} = 16 M_S$

For a dilute solution with $0.25\%$ (mass/mass) concentration, the mass of solute $w_B = 0.25$ g and the mass of solvent $w_S \approx 100$ g.
The molality $m$ is:
$m = \dfrac{w_B \times 1000}{M_B \times w_S} = \dfrac{0.25 \times 1000}{10 M_S \times 100} = \dfrac{0.25}{M_S}$

Given $\Delta T_b = 408 - 400 = 8$ K, we substitute into the boiling point elevation formula:
$8 = i \times (16 M_S) \times \left(\dfrac{0.25}{M_S}\right)$
$8 = 4i \implies i = 2$

The dissociation reaction is $\textbf{B} \rightleftharpoons 2\textbf{C} + 2\textbf{D}$.
The van't Hoff factor $i$ is related to the degree of dissociation $\alpha$ by:
$i = 1 + (n - 1)\alpha$
Here, $n = 4$ (since 1 molecule of $\textbf{B}$ yields 4 particles).
$2 = 1 + (4 - 1)\alpha$
$3\alpha = 1 \implies \alpha = \dfrac{1}{3}$

The mole percent of $\textbf{B}$ that has been dissociated is $\alpha \times 100 = 33.33\%$.

Answer: $33.33$