Answer
$P_2=2.03\times 10^4~Pa$
Work Step by Step
$A_2=2A_1$
Use Fluid Continuity equation to find $v_2$:
$A_1v_1=A_2v_2$
$v_2=v_1({A_1 \over 2A_1})={v_1 \over 2}={2.5~m/s \over 2}=1.25~m/s$
Since we know the pipe is horizontal there, the terms with vertical component will zero out in Bernoulli's Equation, therefore we get:
$P_1+{1 \over 2}\rho v_1^2=P_2+{1 \over 2}\rho v_2^2$
$P_2=P_1+{1 \over 2}\rho v_1^2-{1 \over 2}\rho v_2^2$
$P_2=P_1+{1 \over 2}\rho(v_1^2-v_2^2)$
$P_2=(1.8 \times 10^4~Pa)+{1\over2}(1000~kg/m^3)[(2.5~m/s)^2-(1.25~m/s)^2]=2.03\times10^4~Pa$
