$$V = lwh = (2.13 \text{ m})(1.52 \text{ m})(0.381 \text{ m}) = 1.23 \text{ m}^3$$
The density of water is 1000 kg/m^3, so the mass of the water is:
$$m_w = \rho V = (1000 \text{ kg/m}^3)(1.23 \text{ m}^3) = 1230 \text{ kg}$$
The total weight of the bed is then:
$$W = m_fg + m_ww = (91 \text{ kg})(9.81 \text{ m/s}^2) + (1230 \text{ kg})(9.81 \text{ m/s}^2) = 13000 \text{ N}$$
The pressure exerted on the floor is then:
$$P = \frac{W}{A} = \frac{13000 \text{ N}}{(2.13 \text{ m})(1.52 \text{ m})} = 3900 \text{ Pa}$$
Therefore, the pressure exerted on the floor is 3900 Pa.
