The molal elevation constant for water is 0.52^ ^ C kg mol^ -1 ....

Given:
Mass of urea, $w_2 = 6.0$ g
Molar mass of urea, $M_2 = 60$ g/mol
Mass of water, $w_1 = 200$ g $= 0.2$ kg
Molal elevation constant, $K_b = 0.52^{{}^{\circ}}$C kg mol$^{-1}$

Number of moles of urea is:
$n = \dfrac{w_2}{M_2} = \dfrac{6.0}{60} = 0.1$ mol

Molality of the solution is:
$m = \dfrac{n}{w_1 \text{ (in kg)}} = \dfrac{0.1}{0.2} = 0.5$ mol kg$^{-1}$

Elevation in boiling point is given by:
$\Delta T_b = K_b \times m$
$\Delta T_b = 0.52 \times 0.5 = 0.26^{{}^{\circ}}$C

The boiling point of pure water is $100^{{}^{\circ}}$C.
Therefore, the boiling point of the solution is:
$T_b = 100 + \Delta T_b = 100 + 0.26 = 100.26^{{}^{\circ}}$C

Answer: 100.26^{}^{\circ}C