Answer
$P=126kPa$
Work Step by Step
Choosing $1$ at a depth of $3m$ en $2$ at a depth of $9m$ and knowing that:
$P_{1}=\rho_{fluid}*g*h_{1}$ and $P_{2}=\rho_{fluid}*g*h_{2}$
Assuming that the variation of the density is negligible:
$\rho_{fluid}=\frac{P_{1}}{g*h_{1}}$ and $P_{2}=\frac{P_{1}}{g*h_{1}}*g*h_{2}=\frac{P_{1}*h_{2}}{h_{1}}$
Then:
$P_{2}=\frac{(42kPa)*(9m)}{3m}=126kPa$
