– The density of the fluid $rho = 920 ; frac{text{kg}}{text{m}^3}$

– The pressure at the top of the box $P_1 = 105.0text{ kPa}$

– The atmosphere pressure $P_{text{atm}} = 101.3text{ kPa}$

The pressure at a given depth can be found as follows:

$$
begin{equation*}
P(h) – P_{text{atm}} + rho , g , h
end{equation*}
$$

For the top of the box $h = d , P(h) = P_1$:

$$
begin{equation*}
P_1 = P_{text{atm}} + rho , g , d
end{equation*}
$$

Rearranging for $d$:

$$
begin{equation*}
d = frac{P_1 – P_{text{atm}}}{rho , g}
end{equation*}
$$

Plugging in the numbers we have:

$$
begin{align*}
d &= frac{105.0text{ kPa} – 101.3text{ kPa}}{920 ; frac{text{kg}}{text{m}^3} cdot 9.81 ; frac{text{m}}{text{s}^2}} \
d &approx 0.00041text{ m}
end{align*}
$$

$$
p=p_{atm} + rho g d =2.27 text{ MPa}.
$$

$$
P = P_{atmospheric} + rho gh
$$

$$
h = dfrac{P – P_{atmospheric}}{rho g} = dfrac{1.21 times 10^5 Pa – 1.01 times 10^5 Pa}{(796 kg/m^3)(9.81m/s^2}
$$

$$
h = 2.56m
$$