${P}_{{B}}^{\circ}=15\, {~mm} \,{Hg}$

$\text { and }{X}_{{A}}=0.6,{X}_{{B}}=0.4$

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

$=(0.6 \times 90)+(0.4 \times 15)$

$=54+6=60\, {~mm}$

Now mol fraction of $B$ in the vapour phase

i.e. $Y_{B}=\frac{P_{B}}{P_{T}}=\frac{X_{B} P_{B}^{\circ}}{60}=0.1=1 \times 10^{-1}$

Therefore: $x=1$