The upper-voted answer has the right aim, but it still doesn't give the right estimate. The true source of energy under consideration is indeed the core, not the water, but the calculation in that post is essentially assuming cooling the core from its standing temperature to normal temperature - and as Emilio Pisanty pointed out in the comments, this won't happen as the core is actually its own energy source, capable of maintaining an elevated temperature.
Hence, what you get is effectively a heater that will apply to whatever is applied against it a thermal transfer power equal to the wattage being created by the ongoing fission process within the core material. As such, it is legitimate, as that poster also mentioned, to surmise in theory that an upper limit of megatonnes of total core potential energy is available. In particular, if you have roughly (using figures floating here) 1000 Mg of nuclear fuel that is maybe 5% fissile uranium ($^{235}_{92}\mathrm{U}$), that is 200 Mg of such and this fuel has an energy content of about $86 \times 10^9\ \mathrm{MJ/Mg}$, so the total available energy is on the order of $1.3 \times 10^{11}\ \mathrm{MJ}$, while a megatonne equivalent TNT is roughly $4 \times 10^9\ \mathrm{MJ}$, hence easily tens of megatonnes of available potential fission energy.
But this energy cannot turn into an explosion of that same size under these condition because the core is not releasing that energy fast enough. If it were, it would have already exploded in the manner of a gigantic pure-fission nuclear weapon of that yield. The rate of the fission reaction depends on the composition of the core melt mix, and to get such a reaction would require extreme concentration of the fissile $^{235}_{92}\ \mathrm{U}$ (basically, so that the nuclei are close together and there are few to no obstacles to absorb the neutrons that are needed to propagate the chain reaction), but melting and mixing the material can only serve to dilute it at best. Increasing fissile concentration is the definition of "uranium enrichment" and as we all know, that's HARD! Dumping water on it won't change that. Instead, you a better model would be a a thermal terminal that maintains a constant temperature of 2800 C against anything that hits it, or, at least, something suitably well above the boiling point of water.
Thus, in fact, the question asker is right to imagine this instead as asking for the energy required to vaporize all the water, and this is the maximum energy that can be released in a steam explosion. Energy is contact-transferred - hence once converted to steam, it is very difficult to absorb more from the core.
And this is relatively simple to obtain. With $7000\ \mathrm{m^3}$ of water volume, that's $7000\ \mathrm{Mg}$ of water mass, and the heat of vaporization for water is $2260\ \mathrm{kJ/kg} = 2260\ \mathrm{MJ/Mg}$ (hence my use of megajoules as the unit above), but we also need to take into account the energy to heat the water to the boiling point, which means we should use $4.184\ \mathrm{\frac{kJ}{kg \cdot K}} = 4.184\ \mathrm{\frac{MJ}{Mg \cdot K}}$ times the temperature rise (75 K) which gives $314\ \mathrm{\frac{MJ}{Mg}}$ and hence $2574\ \mathrm{\frac{MJ}{Mg}}$ of total energy to vaporize each megagram (tonne) of water starting at the given staid temperature 25 °C. With 7000 Mg of water, thus, the total potential energy is thus about
$$1.8 \times 10^7\ \mathrm{MJ}$$
maximum possible steam explosion energy. In terms of tonnes equivalent of TNT, it is ~4 kilotonnes TNT equivalent, and so still well below the range given (though also well in excess of the present top answer's figure).