At 363K , the vapour pressure of A is 21kPa and that of B is 18kPa...

Given ${P}_{A}^{0} = 21kPa ⇒ {P}_{B}^{0} = 18kPa$

$\to$ An Ideal solution is prepared by mixing $1$ mole $A$ and $2$ mole $B$.

$\to {X}_{A} = \dfrac{1}{3}$ and ${X}_{B} = \dfrac{2}{3}$

$\to$ According to Raoult's law

${P}_{T} = {X}_{A}{P}_{A}^{0} + {X}_{B}{P}_{B}^{0}$

$⇒ {P}_{T} = \left(\dfrac{1}{3} \times 21\right) + \left(\dfrac{2}{3} \times 18\right)$

$⇒ {P}_{T} = 7 + 12 = 19kPa$